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## Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions

### 2nd PUC Chemistry Solutions NCERT Textbook Questions and Answers

Question 1

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer:

A true solution is a homogenous mixture of two or more substances. The constituent . particle which is in larger amount’’ is called a solvent and that in smaller quantity is called a solute.

Since the solvent ans solute may be either . gaseous, liquid and solid, the number of possible types of binary solutions than can be prepared are given below.

Question 2.

Suppose a solid solution is formed are very small. What kind of solid solution is this likely to be?

Answer:

Solid in solid type. E.g: Copper in gold. This type of solutions are called alloys.

Question 3.

Define the following terms:

(i) Mole fraction

(ii) Molality

(iii) Molarity

(iv) Mass percentage.

Answer:

(i) Mole fraction (X): The mole fraction of any component in a solution is the ratio of the number of moles of that component to the

sum of the number of moles of all the components present in the solution.

For a binary solution containing A and B, Mole fraction of A,

where n_{A} and n_{B} are the numbers of moles of components A and B respectively.

(ii) Molality (m): Molality is the numberof moles of the solute dissolved in 1000 gms (1 kg) of the solvent. It B denoted by ‘m’ mathematically.

(iii) Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm^{3}) of solution. It is denoted by ‘M’ mathematically.

(iv) Mass fraction multiplied by 100 gives mass percentage. E.g.: mass percentage of A

Question 4.

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mLr^{-1}?

Answer:

68% of nitric acid by mass means that

Mass of nitric acid = 68g

Mass of solution = 100g

Molar mass of HN03 = 63g mol^{-1}?

∴ 68 g HNO_{3} = \(\frac{68}{63}\) mole = 1.079 mole

Density of solution = 1.504g mL^{-1}

Question 5.

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL^{-1}, then what shall be the molarity of the solution?

Answer:

10 g glucose is present in 100 g solution, i.e., 90 g of water = 0.090 kg of H_{2}O

Question 6.

How many mL of 0.1 M HCl are required to react completely with 1 g mixture Of Na_{2}CO_{3} and NaHC03 containing equimolar amounts of both?

Answer:

Step 1: To calculate the number of moles of the components in mixture

suppose Na_{2}CO_{3} present in the mixture = X g ,

NaHCO_{3} present in the mixture = (1 – x) g

Molar mass of Na_{2}CO_{3} = 2 x 23+12+3 x 16=106g mol^{-1}

Molar mass of NaHCO_{3}.

= 23 + 1 + 12 + 3 × 16 = 84g mol^{-1}

∴ Moles of Na_{2}CO_{3} in xg = \(\frac{ x }{ 106}\)

Moles of NaHCO_{3} in (1-x) g = \(\frac{1-x}{84}\)

Step 2: To calculate the moles of HC1 required.

Na_{2}CO_{3} + 2 HCl → 2 NaCl + H_{2}O + CO_{2}

NaHCO_{3 }+ HCl → NaCl + H_{2}O + CO_{2}

1 mole of Na_{2}CO_{3} required HCl = 2 moles

0.00526 mole of Na_{2}CO_{3} requires HCl

= 0.00526 × 2 moles = 0.01052

1 mole of NaHCO_{3} required HCl = 1 mole

0.00526 mole of NaHCO_{3} required HCl

= 0.00526 mole

∴Total HCl required = 0.01052 + 0.00526

=0.01578 moles

Step3: To calculate volume of 0.1M HCl

0.1 mole of 0.1 M HCl are present in 1000 mL of HCl

0.01578 mole of 0.1 M HCl will be present in HCl = \(\frac{1000}{0.1}\) x 0.01578

= 157.8 ml.

Question 7.

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Answer:

300 g of 25% solutions contains solute = 75g

400 g of 40% solution contains solute = 160 g.

Total solute = 160 + 75 = 235 g

Total solution=300 + 400 = 700 g

% of solute in final solution = \(\frac{235}{700}\) × 100 = 33.5%

% of water in the final solution = 100 – 33.5 = 66.5%

Question 8.

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_{2}H_{6}O_{2}) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^{-1}, then what shall be the molarity of the solution?

Answer:

Mass of the solute, C_{2}H_{4}(OH)_{2} = 222.6g

Molar mass of C_{2}H_{4} (OH)_{2} = 62 g mol^{-1}

Mass of the solvent = 200 g = 0.200 kg

Total mass of the solution = 422.6g

Volume of the solutions =

Question 9.

A sample of drinking water was found to be severely contaminated with chloroform (CHCl_{3}) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

(i) express this in percent by mass

(ii) determine the molality of chloroform in the water sample.

Answer:

15 ppm means 15 parts in million (10^{6}) parts by mass in solution

∴ % by mass = \(\frac{15}{10^{6}} \times\) × 100=15 × 10^{-4}

Taking 15 g chloroform in 10^{6}g

Molar mass of CHCl_{3} = 12 + 1 + 3 × 35.5 = 119.5 g mol^{-1}

Molality = \(\frac{15 / 119.5}{10^{6}}\) × 1000= 1.25 × 10^{-4} m

Question 10.

What role does the molecular interaction play in a solution of alcohol and water?

Answer:

Alcohols dissolve in water due to the formation of intermolecular H-bonding with water.

Question 11.

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer:

Dissolution of gas in liquid in an exothermic process (Gas + solvent ⇌ solution + Heat). As the temperature is increased, the equilibrium shifts backward.

Question 12.

State Henry’s law and mention some important applications?

Answer:

Henry’s law: According to this law, ‘The mass of a gas dissolved per unit volume of a solvent at constant temperature, is proportional to the pressure of the gas with which the solvent is in equilibrium’.

Let in unit volume of solvent, mass of the gas dissolved is m and equilibrium pressure be P, then m α P or m = KP, where K is a constant. We can understand Henry’s law by taking example of soda water bottle. Soda water contains carbon dioxide dissolved in water under pressure.

Applications of Henry’s law:

1. In the production of carbonated beverages: To increase the solubility of CO_{2} in soft drinks, soda water, bear etc. the bottles are sealed at high pressure.

2. In exchange of gases in the blood: The partial pressure of O_{2} is high inhaled air, in lungs it combines with hemoglobin to form oxyhemoglobin. In tissues, the partial pressure of oxygen is comparatively low therefore oxyhemoglobin releases oxygen in order to carry out cellular activities.

3. In deep-sea diving: Deep-sea divers depend upon compressed air for breathing at high pressure underwater. The compressed air contains N_{2} in addition to O_{2}, which are not very soluble in blood at normal pressure. However, at great depths when the diver breathes in compressed air from the supply tank, more N_{2} dissolve in the blood and in other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. When the divers come towards the surface at atmospheric pressure, this dissolve nitrogen bubbles out of the blood. These bubbles restrict blood flow, affect the transmission of nerve impulses. This causes a disease called bends or decompression sickness. To avoid bends, as well as toxic effects of high concentration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% He, 56.2% N_{2} and 32.1% O_{2}).

4. At high altitudes: At high altitudes the partial pressure of O_{2} is less than that at the ground level. This result in low concentration of oxygen in the blood and tissues of the people living at high altitudes or climbers. The low blood oxygen causes climbers to become weak and unable to think clearly known as anoxia.

5. Aquatic life : The dissolution of oxygen (from air) in water helps in the existence of aquatic life in various water bodies like : Lake, rivers and sea.

Question 13.

The partial pressure of ethane over a solution containing 6.56 x 10^{-3} g of ethane is 1 bar. If the solution contains 5.00 x 10^{-2} g of ethane, then what shall be the partial pressure of the gas?

Answer:

Applying the relationship m = K_{H} x p

In the first case, 6.56 x 10^{2} g bar^{-1}

In the second case, 5.00 x 10^{-2} g = (6.56 × 10^{-2} gbar^{-1}) × p

Question 14.

What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_{mix}H related to positive and negative deviations from Raoult’s law?

Answer:

In +ve deviation, A-B interactions are weaker than those between A-A or B-B. In such molecules, A or B will find it easier to escape than in pure state. This increases vapour .pressure. In case of -ve deviation, A-B interaction = A-A or B-B. This leads to decrease in vapour pressure.

- In +ve deviation, Δ
_{mix}H is + ve - In -ve deviation, Δ
_{mix}H is – ve

Question 15.

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer:

Vapour pressure of pure water at boiling

point (P°) = 1 atm = 1.013 bar

vapour pressure of solution (p_{s}) = 1.004 bar

Mass of solute = (w_{2}) = 2g

Mass of solution = 100 g

Mass of solvent = 98g

Applying Roault’s law for dilute solution (being 2%)

Question 16.

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Answer:

Molar mass of heptane (C_{7} H_{16}) = 100 g mol^{-1}

Molar mass of octane (C_{8}H_{18}) = 114 g mol^{-1}

x (octane) = 1 – 0.456 = 0.544

p (heptane) = 0.456 x 105.2 kPa = 47.97 kPa

p (octane) = 0.544 x 46.8 kPa = 25.46 kPa

p total = 47.97 + 25.46 = 73.43 kPa

Question 17.

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer:

1 molal solution means 1 mol of the solute in 1 kg of solvent (water)

or p_{s} = 12.08 kPa

Question 18.

Calculate the mass of a non-volatile solute (molar mass 40 g mol^{-1}) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Answer:

Ps = 80% of p° = 0.80 p° solute = \(\frac{w}{40} r\) mol solvent (octane) =

Question 19.

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

(i) molar mass of the solute

(ii) vapour pressure of water at 298 K.

Answer:

Suppose the molar mass of the solute = Mg mol^{-1}

n_{2} (solute) = \(\frac{30}{M}\) moles

After adding 18 g of water,

n(H_{2}O)i.e, n_{1} = 6 moles

Dividing equation (i) by equation (ii), we get

(ii) putting M = 23 in equation (i), we get

Question 20.

A 5% solution (by mass) of cane sugar in water has freezing point of 271 A. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer:

Molality of sugar solution =

∆T_{f }for sugar solution = 273.15 – 271 = 2.15^{0}

∆T_{f} = K_{f} × m ∴ K_{f} = 2.15/0.146

Molality of glucose solution =

Freezing point of glucose solution = 273.15 – 4.09 = 269.06 k

Question 21.

Two elements A and B form compounds having formula AB_{2} and AB_{4}. When dissolved in 20 g of benzene (C_{6}H_{6}), 1 g of AB_{2} lowers the freezing point by 2.3 K whereas 1.0 g of AB_{4} lowers it by 1.3 K. Tin molar depression – constant for benzene is 5.1 K kg mol^{-1}. Calculate atomic masses of A and B.

Answer:

Suppose atomic masses of A and B are ‘a’ and ‘b’ respectively. Then

Molar mass of AB_{2} = a + 2b = 110.87 g mol^{-1}

Molar mass of AB_{4} = a + 4b =196.15 g mol^{-1}

Equation (ii) – Equation (i) gives

2b = 85.28 orb = 42.64

substituting in equation (i) we get

a + 2 × 42.64 = 110.87 or a = 25.59

Thus atomic mass A = 25.59 u

Atomic mass of B = 4.64 u.

Question 22.

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Answer:

π = CRT

∴ In the first case,

4.98 = \(\frac{36}{180}\) × R × 300 = 60R

In the second case, 1.52 = C × R × 300

Dividing (ii) by (i), we get C = 0.061 M.

Question 23.

Suggest the most important type of intermolecular attractive interaction in the following pairs.

- n-hexane and n-octane
- I
_{2}and CCl_{4} - NaClO
_{4}and water - methanol and acetone
- acetonitrile (CH
_{3}CN) and acetone (C_{3}H_{6}O).

Answer:

- Both are non-polar. Hence, intermolecular interactions in them will be London/ dispersion forces (discussed in class XI)
- Both are non-polar. Hence, intermolecular interactions in them will be London/ dispersion forces (discussed in class XI)
- NaClO
_{4}gives Na+ and ClO_{4}^{–}ions in the solution while water is a polar molecule. Hence, intermolecular interactions in them will be ion-dipole interactions. - Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.
- Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.

Question 24.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH_{3}OH, CH_{3}CN.

Answer:

(i) Cyclohexane and n-octane both are non-polar. Hence they mix completely in all proportions.

(ii) KCl is an ionic compound while n-octane is nonpolar. Hence, KCl will not dissolve at all in n-octane.

(iii) CH_{3}OH and CH_{3}CN both are polar but CH_{3}CN is less polar than CH_{3}OH. As the solvent is non-polar, CH_{3}CN will dissolve more than CH_{3}OH is n-octane.

Thus the order of solubility will be KCl< CH_{3}OH < CH_{3}CN < Cyclohexane.

Question 25.

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i) phenol

(ii) toluene

(iii) formic acid

(iv) ethylene glycol

(v) chloroform

(vi) pentanol.

Answer:

(i) Partially soluble (because phenol has

polar-OH group but aromatic phenyl, C_{6}H_{5 }– group)

(ii) Insoluble because toluene is nonpolar while water is polar

(iii) Highly soluble because formic acid can form hydrogen bonds with water.

(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water

(v) Insoluble chloroform is an organic liquid

(vi) Partially soluble because-OH the group is polar but the large hydrocarbon part (C_{5}H_{11}) is nonpolar.

Question 26.

If the density of some lake water is 1.25g mL^{-1} and contains 92 g of Na^{+} ions per kg of water, calculate the molality of Na+ ions in the lake.

Answer:

Number of moles in 92 g of Na^{+} ions

As these are present in 1 kg of water, by definition, molality = 4m.

Question 27

If the solubility product of CuS is 6 × 10^{-16}, calculate the maximum molarity of CuS in aqueous solution.

Answer:

Maximum molarity of CuS in aqueous solution = solubility of CuS in mol L^{-1}

If S in the solubility of CuS in mol L^{-1} , then

Question 28.

Calculate the mass percentage of aspirin (C_{9}H_{8}O_{4}) in acetonitrile (CH_{3}CN) when 6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.

Answer:

Question 29.

Nalorphene (C_{19}H_{21}NO_{3}), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10^{-3} m aqueous solution required for the above dose.

Answer:

1.5 × 10^{-3} m solution means that 1.5 × 10^{-3} mole of nalorphine is dissolved in I kg of water.

Molar mass of C_{19}H_{21}NO_{3} = 19 × 12 + 21 + 14 + 48 = 3119 mol^{-1}

∴ 1.5 × 10^{-3} mole of C_{19}H_{21}NO_{3};

= 1.5 × 10^{-3} × 3119 = 0.467 g = 467 mg

∴ Mass of solution = 1000 g + 0.467 g = 1000.467g

Thus, for 467 mg of nalorphene, solution required =1000.467 g for 1.5 mg of nalorphene, solution required =\(\frac{1000.467}{467} \times 1.5 = 3.219\)

Question 30.

Calculate the amount of benzoic acid (C_{6}H_{5}COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Answer:

0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L

i.e. 1000 mL of the solution.

Molar mass of benzoic acid (C_{6}H_{5}COOH) = 72 + 5 + 12 + 32 + 1 =122 g mol^{-1}

∴ 0.15 mole of benzoic acid = 0.15 × 122g= 18.39

Thus, 1000 mL of the solution contain benzoic acid = 18.39

∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250= 4.575g

Question 31.

The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic acid increases in the order given above. Explain briefly.

Answer:

The depression in freezing points are in the order:

acetic acid < trichloroacetic acid

Fluorine, being most electronegative, has the highest electron-withdrawing inductive effect. Consequently, trifluoroacetic acid is the Strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to minimum extent to give ions in their solutions in water. Greater the ions produced, greater is the depression in freezing point. Hence the depression in freezing point is maximum for trifluoroacetic acid and minimum for acetic acid.

Question 32.

Calculate the depression in the freezing point of water when 10 g of CH_{3}CH_{2} CHCICOOH is added to 250 g of water. K_{a} = 1.4 × 10^{-3}, K_{f}= 1.86 K kg mol^{-1}.

Answer:

Molar mass of CH_{3}CH_{2}CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol^{-1}

If α is the degree of dissociation of CH_{3}CH_{2}CHCICOOH, then CH_{3}CH_{2}CHCLCOOH □ CH3CHXHCICOO^{–} + H +

To calculate vant’s Hoff factor:

CH_{3}CH_{2}CHCICOOH □ CH3CH_{2}HCICOO^{–} + H^{+}

Question 33.

19.5 g of CH_{2}FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoro acetic acid. K_{f} of water is 1.86 K kg mol^{-1}.

Answer:

Here, w_{2} = 19.5 g, w_{1}, = 500g, K_{f }= 1.86 K kg mol^{-1}, (AT_{f}) obs = 1.0°

M_{2} (Calculated) for CH_{2} FCOOH = 14+19+45 = 78 g mol^{-1}

Vant Hoff factor (i) =

calculation of dissociation constant. Suppose degree of dissociation at the given concentration is a.

Then

Taking volume of the solutions as 500ml,

Question 34.

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer:

Here, P° = 17.535 mm, w_{2} = 25g,

w_{1} = 450g

For solute (glucose, C_{6}H_{12}O_{6}, M_{2} = 180 g mol^{-1}

For solvent (H_{2}O), M_{1} = 18g mol^{-1}

Applying Raoult’s law,

substituting the given values, we get

substituting the given values, we get

Question 35.

Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10s mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer:

Here, K_{H} = 4.27 × 10^{5} mm

p = 760 mm

Applying Henry’s law

P = K_{H} x

i.e, mole fraction of methane in benzene = 1.78 x 10^{-3}

Question 36.

100 g of liquid A (molar mass 140 g mol^{-1}) was dissolved in 1000 g of liquid B (molar mass 180 g mol^{-1}). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Answer:

Number of moles of a liquid

Number of moles of a liquid B (Solvent)

Mole fraction of B in the solution (x_{A})

Mole fraction of B in the solution (x_{B})

= 1-0.114 = 0.886

Also, given p^{0}_{B} = 500 Torr

Applying Raoult’s law,

P_{A} = x_{A}P°_{A} = 0.114 × P°_{A }

P_{B}= x_{B} p°_{B} = 0.886 × 500 = 443 Torr

p Total = p_{A }+ p_{B}

475 = 0.114 P°_{A} +443 or

Substituting this value in equation (i), we get p_{A }= 0.114x 280.7 Torr = 32 Torr

Question 37.

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50. 71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer:

Molar mass of benzene (C_{6}H_{6} = 78 g mol^{-1}

Molar mass of toluene (C_{6}H_{5}CH_{3}) = 92 g mol^{-1}

∴ Number of moles in 80 g of benzene

Number of moles in 100 g of toluene

In the solution, mole fraction of benzene

=0.486

mole fraction of toluene = 1-0,486 = 0.514

p° Benzene = 50.71 mm, p° Toluene = 32.06mm

Applying Raoult’s law,

p Benzene= x_{Tolucno} × p°_{Tolucne} =0.514 × 32.06 mm = 16.48mm

Mole fraction of benzene in vapour phase

Question 38.

The air is a mixture of a number of gases. The major components are oxygen and, nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10^{7} mm and 6.51 × 10^{7} mm respectively, calculate the composition of these gases in water.

Answer:

Total pressure of air in equilibrium with water = 10 atm

As air contains 20% oxygen and 79% nitrogen by volumes

partial pressure of oxygen :

(P ) = \(\frac{20}{100}\) x 10 atm = 2atm

= 2 x 760 mm = 1520 mm

partial pressure of nitrogen \(P_{N_{2}}\)

= \(\frac{79}{100}\) × 10 atm = 7.9 atm = 7.9 x 760 mm

= 6004 mm

K_{H}(O_{2}) = 3.30 x 10^{7}mm,K_{H} (N_{2}) = 6.51 × 10^{7} mm

Applying Henry’s law

Question 39.

Determine the amount of CaCl (i= 2.47) dissolved in 2.51itre of water such that its osmotic pressure is 0.75 atm at 27° C.

Answer:

Molar mass of CaCl_{2}= 40 + 2x 35.5 = 111 gmol^{-1}

Amount dissolved = 0.0308 xlllg = 3.42g

Question 40.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K_{2}SO_{4}in 2 litre of water at 25° C, assuming that it is completely dissociated.

Answer:

K_{2}SO_{4} dissolved = 0.025g

volume of solution = 2L

T = 250°C = 298 K

Molar mass of K_{2}SO_{4} =

2×39 + 32 + 4 × 16= 174gmol^{-1}

As, K_{2}SO_{4} dissociates completely as K_{2}SO_{4} → 2K+ SO_{4}^{2-}

i.e., ions produced = 3 ∴ i =3

### 2nd PUC Chemistry Solutions Additional Questions and Answers

Question 1.

How does molarity of a solution change with temperature?

Answer:

Molarity decreases with increase in temperature because volume of solution increases with increase in temperature.

Question 2.

State Raoult’s law.

Ans:

It states that for a solution of volatile liquids its vapour pressure of each component is directly proportional to their mole fraction in the solution.

The relative lowering of vapour pressure is equal to mole fraction of solute in case of non volatile solute.

Question 3.

Why is liquid ammonia bottle first cooled in ice before opening it?

Answer:

At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out.

Question 4.

Sodium chloride and calcium chloride are used clear show from the roads. Why?

Answer:

Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature.

Question 5.

Two liquids A and B boil at 145°C and 190°C

respectively. Which of them has a higher vapour pressure of 80°C? (CBSE 2006)

Answer:

A being more volatile will have higher vapour pressure at 80°C.

Question 6.

Calculate the molality of sulphuric acid solution in which the mole fraction is 0.85.

Answer:

Question 7.

The molar freezing point depression constant for benzene is 4.90 kg mol’1. Selenium exists as polymer. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower for pure benzene. Decide the molecular formula of selenium (At Wt. of selenium is 78.8 g mol^{-1}). (CBSE 2002)

Answer:

Question 8.

CCI_{4} and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. (CBSE 2003,2001)

Answer:

CCI_{4} is a non-polar covalent compound.

Water is a polar compound. CCI_{4} can neither form H^{–} bonds with water molecules nor can it break H^{–} bonds between water molecules. Therefore, it is insoluble in water.

Ethanol is a polar compound and can form H’ bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions.

Question 9.

The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm^{3} )

Answer:

Let the solution be 1 litre or 1000 cm^{3}

∴ Number of moles of H_{2}SO_{4}= 1.35

Wt. of solution = 1000 x 1.02 = 1020 g

Wt. of sulphuric acid = 1.35 x 98= 132.3g :

Wt. of water = 1020 – 132.3 = 887.79

Molality of H_{2}SO_{4}= ppp x 1000 = 1.52 m

From (i) M, = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15- 110.82 = B2 85.33 = B2 B = 42.665

Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82=Atomic mass of A+85.33 Atomic mass of A = 110.82 – 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669

Question 10.

Two elements A and B form compounds having molecular formula AB2 and AB4. When dissolved 20 g of C6H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB„4 lower it by 1.3 K. The molar depression constant for benzene is 5.1 kg mol*1. Calculate the mass of A and B. (CBSE2004)

Answer:

Let the molar mass of AB_{2} and AB_{4} be M_{1}and M_{2}

Then, for AB_{2},

From (I) M_{1} = 110.82, from (ii) M_{1} = 196.15

AB_{4}– AB_{2}= B_{2}

196.15 – 110.82 =B_{2}

85.33 = B_{2}

B = 42.665

Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B

110.82 = Atomic mass of A+ 85.33

∴ Atomic mass of A = 110.82-85.33 = 25.49

Atomic mass of A = 25 .499

Atomic mass of B = 42.669