2 g of benzoic acid dissolved in 25 g of C 6 H 6 show a depression in freezing point equal to 1 . 62 K . Molal depression constant of C 6 H 6 is 4 . 9 K mol - 1 Kg . What is the percentage association of acid if it forms double molecules in solution?

Given, w = 2 g , W = 25 g , ΔT = 1 . 62 , K ' f = 4 . 9 ∵ ΔT = 1000 × K ' f × w m × W 1 . 62 = 1000 × 4 . 9 × 2 25 × m ∴ m exp = 241 . 98 g mol - 1 nC 6 H 5 COOH ⇌ C 6 H 5 COOH n Before association 1 0 After association ( 1 - α ) α n ∴ Total number of mole at equilibrium = 1 - α + α n m N m exp = 1 - α + α n For dimer formation n = 2 122 . 0 241 . 98 = 1 - α + α 2 m N = 122 . 0 for C 6 H 5 COOH or 1 - α 2 = 0 . 504 ∴ α = 0 . 992 or 99 . 2 %

x g of a non-electrolytic compound (molar mass = 200 ) are dissolved in 1 . 0 L of 0 . 05 M NaCl aqueous solution. The osmotic pressure of this solution is found to be 4 . 92 atm at 27 ° C . Calculate the value of x . Assume complete dissociation of NaCl and ideal behaviour of the solution ( R = 0 . 0821 L atm K - 1 mol - 1 ) .

For the non-electrolytic compound: w = x g , molar mass = 200 , V = 1 L , T = 27 ° C = 300 K , R = 0 . 0821 L atm K - 1 mol - 1 Osmotic pressure =? Moles of solute n 1 = x 200 πV = n RT π 1 = n 1 RT V = x 200 × 0 . 0821 × 300 1 = 0 . 1231 atm For NaCl solution NaCl ⇌ Na + + Cl - Initially 1 mole 0 0 At equilibrium - 1 mole 1 mole i = No . of moles in solution No . of moles added = 1 + 1 1 = 2 Molarity of NaCl solution = 0 . 05 M ∴ Moles of NaCl n 2 = 0 . 05 , V = 1 L , T = 300 K , π 2 = ? According to the modified equation of osmotic pressure πV = i nRT π 2 × 1 = 2 × 0 . 0821 × 300 = 2 . 463 atm Given that the osmotic pressure of the solution of non-electrolyte compound in 0 . 05 M NaCl at 27 ° C is 4 . 92 atm . This must be equal to the sum of osmotic pressure due to the compound π 1 and due to 0 . 05 M NaCl π 2 . Therefore, π 1 + π 2 = 4 . 92 or 0 . 1231 x + 2 . 463 = 4 . 92 x = 4 . 92 - 2 . 463 0 . 1231 = 19 . 96 g

The degree of dissociation of Ca NO 3 2 in a dilute aqueous solution containing 7 . 0 g of the salt per 100 g of water at 100 ° C is 70 per cent. If the vapour pressure of water at 100 ° C is 760 mm , calculate the vapour pressure of the solution.

p 0 - p s p 0 = ( i ) n A n A + n B p 0 = 760 mm n A = 7 . 0 164 n B = 100 18 i = 1 + 2 α Ca NO 3 2 ⇌ Ca 2 + + 2 NO 3 - Before dissociation 1 0 0 After dissociation 1 - α α 2 α ∴ Total mole at equilibrium = ( 1 + 2 α ) ( ∵ α = 0 . 7 ) = ( 1 + 2 × 0 . 7 ) = 2 . 4 For Ca NO 3 2 : m n m exp = 1 + 2 α ∴ m exp = m n ( 1 + 2 × 0 . 7 ) = 164 2 . 4 = 68 . 33 Also at 100 ° C , P ° H 2 O = 760 mm , w = 7 g , W = 100 g Now P o - P s P s = 7 × 18 68 . 33 × 100 = 0 . 0184 or P o P s - 1 = 0 . 0184 ∴ P s = 760 1 . 0184 = 746 . 27 mm

A solution of a non-volatile solute in water freezes at - 0 . 30 ° C . The vapour pressure of pure water at 298 K is 23 . 51 mm Hg and K f for water is 1 . 86 degree / molal . Calculate the vapour pressure of this solution at 298 K .

We have, ΔT = K f × molality Also from Raoult's law, P ° - P s P s = w × M m × W = w × 1000 × M m × W × 1000 P ° - P s P s = molality × M 1000 P ° - P s P s = ΔT K f × M 1000 P ° - P s P s = ΔT K f × M 1000 Given, P ° = 23 . 51 mm of Hg ; ΔT = 0 . 3 K f = 1 . 86 K molality ; - 1 M = 18 ∴ 23 . 51 - P s P = 0 . 3 1 . 86 × 18 1000 ∴ P s = 23 . 44 mm Hg

MEDIUM

12th ICSE

IMPORTANT

Earn 100

A $1.2 %$ solution of $NaCl$ is isotonic with $7.2 %$ solution of glucose. Calculate the van't Hoff factor for $NaCl$ .

Important Questions on Solutions

HARD

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 8

2 g

of benzoic acid dissolved in

25 g

C_{6} H_{6}

show a depression in freezing point equal to

1.62 K

. Molal depression constant of

C_{6} H_{6}

4.9 K {mol}^{- 1} Kg

. What is the percentage association of acid if it forms double molecules in solution?

HARD

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 9

x g

of a non-electrolytic compound (molar mass

= 200

) are dissolved in

1.0 L

0.05 M

NaCl

aqueous solution. The osmotic pressure of this solution is found to be

4.92 atm

27 ° C

. Calculate the value of

x

. Assume complete dissociation of

NaCl

and ideal behaviour of the solution

(R = 0.0821 L atm K^{- 1} {mol}^{- 1})

MEDIUM

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 10

The degree of dissociation of

Ca {({NO}_{3})}_{2}

in a dilute aqueous solution containing

7.0 g

of the salt per

100 g

of water at

100 ° C

70

per cent. If the vapour pressure of water at

100 ° C

760 mm

, calculate the vapour pressure of the solution.

MEDIUM

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 11

Calculate the boiling point of a solution containing

0.61 g

of benzoic acid in

50 g

of carbon disulphide assuming

84 %

dimerisation of the acid. The boiling point and

k_{b}

{CS}_{2}

are

46.2 ° C

and

2.3 K kg {mol}^{- 1}

MEDIUM

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 12

How many grams of

KCl

should be added to

1.00 kilogram

of water to lower its freezing point to

- 8.0 ° C

K_{f} (H_{2} O) = 1.88 ° C kg {mol}^{- 1}

(At. masses :

K = 39, Cl = 35.5

MEDIUM

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 13

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure of

2.0 atm

300 K

MEDIUM

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>NUMERICAL PROBLEMS>Q 14

A solution of a non-volatile solute in water freezes at

- 0.30 ° C

. The vapour pressure of pure water at

298 K

23.51 mm Hg

and

K_{f}

for water is

1.86 degree / molal .

Calculate the vapour pressure of this solution at

298 K

MEDIUM

12th ICSE

IMPORTANT

ICSE Chemistry>Chapter 2 - Solutions>QUESTIONS FROM ISC EXAMINATION PAPERS>Q 1

The freezing point of nitrobenzene is

278.8 K

. A

0.25

molal solution of a substance (mol. wt.

= 120

) in nitrobenzene has a freezing point of

276.8 K

. Calculate the molal depression constant of nitrobenzene.

A 1.2% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the van't Hoff factor for NaCl.

Important Questions on Solutions

Learn and Prepare for any exam you want !

A $1.2 %$ solution of $NaCl$ is isotonic with $7.2 %$ solution of glucose. Calculate the van't Hoff factor for $NaCl$ .