EASY

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When super cooled liquid starts freezing , the temperature

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Important Questions on Solutions

EASY

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

Of the following

0.10 m

aqueous solutions, which one will exhibit the largest freezing point depression?

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

For

1 molal

aqueous solution of the following compounds, which one will show the highest freezing point?

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

Molal depression constant for a solvent is

4.0 K k g m o l^{- 1} .

The depression in the freezing point of the solvent for

0.03 m o l k g^{- 1}

solution of

K_{2} S O_{4}

is:
(Assume complete dissociation of the electrolyte)

EASY

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

The freezing point depression constant

(K_{f})

of benzene is

5.12 K kg {mol}^{- 1}

. The freezing point depression for the solution of molality

0.078 m

containing a non-electrolyte solute in benzene is (rounded off upto two decimal places):

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

The elevation in boiling point for

1

molal solution of glucose is

2 K

. The depression in freezing point for

2

molal solution of glucose in the same solvent is

2 K .

The relation between

K_{b}

and

K_{f}

is:

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

When the same amount of the solute

' P'

and

' Q'

are separately dissolved in

500 g

water, the

Δ

T_{f}

values are

0.15 K

and

0.30 K

respectively. If the molecular weight of

{}^{'}P^{'}

80 g {mol}^{- 1}

, then the molecular weight of

{}^{'}Q^{'}

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

Pure water freezes at

273 K

and

1 bar

pressure. The addition of

34.5 g

of ethanol to

500 g

of water changes the freezing point of the solution. Use the freezing point depression constant of water as

2 K kg {mol}^{- 1}

. The figures shown below represent the plots of vapour pressure (

V . P .

) versus temperature (

T

). [Molecular weight of ethanol is 46 g

46 g {mol}^{- 1}

]

Among the following, the option representing a change in the freezing point is

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

The mass of sucrose in grams (molar mass of sucrose is

(

342 g {mol}^{- 1}

)

required to be dissolved in

100 g

of water, in order to produce a solution with

104.74 ° C

as the difference between the boiling point and freezing point is

(K_{f (H_{2} O)} = 1.86 K kg {mol}^{- 1}

and

K_{b (H_{2} O)} = 0.51 K kg {mol}^{- 1})

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

A solution containing

62 g

ethylene glycol in

250 g

water is cooled to

- 10^{o} C

. If

K_{f}

for water is

1.86 K kg {mol}^{- 1}

, the amount of water (in

g

) separated as ice is:

HARD

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

5 g

{Na}_{2} {SO}_{4}

was dissolved in

x g

H_{2} O

. The change in freezing point was found to be

3 {.82}^{o} C

. If

{Na}_{2} {SO}_{4}

81.5 %

ionised, the value of

x

(

K_{f}

for water

= 1 {.86}^{o} C kg {mol}^{- 1}

) is approximately:

(molar mass of

S = 32 g {mol}^{- 1}

and that of

Na = 23 g {mol}^{- 1}

)

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

Which of the following

0.1 M

aqueous solution will have lowest freezing point?

HARD

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

Molecules of benzoic acid

(C_{6} H_{5} COOH)

dimerise in

30 g

of benzene. '

w

g

of benzoic acid shows a depression in freezing point equal to

2 K

. If the percentage association of the acid to form dimer in the solution is

80

, then

w

is:

(

Given that

K_{f} = 5 K kg {mol}^{- 1}

, molar mass of benzoic acid

= 122 {gmol}^{- 1})

HARD

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

When

2.44

grams of benzoic acid

(C_{6} H_{5} COOH)

dissolved in

25

grams of benzene, it shows depression of freezing point equal to

2.2 K

. Molal depression constant of benzene is

5.0 K kg {mol}^{- 1}

. What is the percentage association of acid, if it forms dimer in solution?

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

A non-volatile solute,

{}^{'}A^{'}

tetramerises in water to the extent of

80 %

2.5 g

{}^{'}A^{'}

100 g

of water, lowers the freezing point by

0.3 ° C

. the molar mass of

A

g {mol}^{- 1}

is (

K_{f}

for water

= 1.86 K kg {mol}^{- 1}

)

HARD

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

The freezing point depression of

0.001 M

A_{x} B_{y} [Fe {(CN)}_{6}]

5.58 \times 10^{- 3} K

. If the oxidation state of

Fe

+ 2

and

K_{f} = 1.86 K {kgmol}^{- 1}

, then the total number of possibilities for different types of

A

and

B

cations are

EASY

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

K_{f}

(water)

= 1.86 K kg {mol}^{- 1}

. The temperature at which ice begins to separate from a mixture of

10

mass

%

ethylene glycol is

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

When

36 g

of a non-volatile, non-electrolytic solute have the empirical formula

{CH}_{2} O

is dissolved in

1.2 kg

of water, the solution freezes at

- 0.93 ° C

. The molecular formula of the solute is (

K_{f}

of water

= 1.86 K kg {mol}^{- 1}

)

EASY

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

If molality of the dilute solution is doubled, the value of molal depression constant

(K_{f})

will be

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

The freezing point of benzene decreases by ${0.45}^{°} C$ on adding $0.2 g$ of acetic acid to $20 g$ of benzene. If acetic acid associates to form a dimer in benzene, then what is the percentage association of acetic acid in benzene?

$(K_{f} for benzene = 5 .12 K kg {mol}^{- 1})$

MEDIUM

Chemistry>Physical Chemistry>Solutions>Depression of Freezing Point

The freezing point of a

4 %

aqueous solution of

X

is equal to the freezing point of a

12 %

aqueous solution of

Y

. If the molecular weight of

X

A

, then the molecular weight of

Y

will be

When super cooled liquid starts freezing , the temperature

Important Questions on Solutions

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