A aqueous solution consisting 0-5 g KCl in 100 mL water was found to freeze at -0-24-C - Find the van-t Hoff factor and degree of dissociation of the solute respectively- Given -Kfwater- 1-86 K kg mol-1-water-1 g mL-1

Since , ρ_water=1 g mL^ 1 Weight of water (W)=100 g Calculating Observed mol. mass of nbsp;KCl (M_B) nbsp; Δ T_f=K_f× m Here m is molality M_B= 1000 × K_ ...

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Standard XII

Chemistry

Van't Hoff Factor

A aqueous sol...

Question

A aqueous solution consisting $0.5 g K C l$ in $100 m L$ water was found to freeze at $- {0.24}^{\circ} C$ . Find the van't Hoff factor and degree of dissociation of the solute respectively.
Given :
$(K_{f})_{water} = 1.86 K k g m o l^{- 1} ρ_{w a t e r} = 1 g m L^{- 1}$

1.92 and 0.92

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1.79 and 0.79

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2 and 0.63

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1.5 and 0.54

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Solution

The correct option is A $1.92 and 0.92$
Since ,
$ρ_{w a t e r} = 1 g m L^{- 1} Weight of water (W) = 100 g$

$Calculating Observed mol. mass of K C l (M_{B}) Δ T_{f} = K_{f} \times m Here m is molality M_{B} = \frac{1000 \times K_{f} \times w}{Δ T \times W}$
where,
$K_{f} = 1.86 K k g m o l^{- 1}$ , $weight of solute (w) = 0.5 g, W = 100 g, Δ T_{f} = (0 - (- 0.24)) K = + 0.24 K$

So,
$M_{B} = \frac{1000 \times 1.86 \times 0.5}{0.24 \times 100} = 38.75 g m o l^{- 1}$

$Theoretical molecular mass of K C l = 39 + 35.5 = 74.5 g m o l^{- 1}$

$van't Hoff factor = \frac{Theoretical molecular mass}{Observed molecular mass} i = \frac{74.5}{38.75} = 1.92$
Option (b) is correct
For dissociation
$α = \frac{i - 1}{n - 1}$

Aqueous solution of $K C l$ dissociates into two ions

$∴ n = 2$
$\Rightarrow α = \frac{1.92 - 1}{2 - 1} α = 0.92$
Option (a) is correct.

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Similar questions

Q. A aqueous solution consisting

0.5 g K C l

100 m L

water was found to freeze at

- {0.24}^{\circ} C

.
Given :

(K_{f})_{water} = 1.86 K k g m o l^{- 1} ρ_{w a t e r} = 1 g m L^{- 1}

Choose the correct statement(s):

Van't Hoff factor, i, of a 0.5% (w/w) aqueous solution of KCl which freezes at

- {0.24}^{o} C

is:

[K_{f}

of water = 1.86 K kg mol

^{- 1}

, Mol.wt.of KCl = 74.5]

Q. Van't Hoff factor, (i), of a

0.5 %

(w/W) aqueous solution of

K C l

which freezes at

- {0.24}^{o}

C is:

(

K_{f}

of water

= 1.86

K kg

m o l^{- 1}

, molecular weight of

K C l

= 74.5

)

0.6 m l

of glacial acetic acid with density

1.06 g / m l

is dissolved in

1 k g

water and the solution froze at

- {0.0205}^{\circ} C

. Calculate van't Hoff factor:

K_{f}

for water is

1.86 K k g m o l^{- 1}

Q. A solution which has

10 g

of solvent and

0.20 g

of solute (mol. mass 122 g/mol) freezes at

{0.5}^{\circ} C

below the freezing point of pure solvent. If the solute exists as dimer in the solvent, choose the correct option :

(K_{f})_{solvent} = 5.12 K k g^{- 1} m o l^{- 1}

Here,

i = Van't hoff factor β = Degree of association

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Van't Hoff Factor

Standard XII Chemistry

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A aqueous solution consisting 0.5 g KCl in 100 mL water was found to freeze at −0.24∘C . Find the van't Hoff factor and degree of dissociation of the solute respectively. Given : (Kf)water=1.86K kg mol−1ρwater=1 g mL−1

A aqueous solution consisting $0.5 g K C l$ in $100 m L$ water was found to freeze at $- {0.24}^{\circ} C$ . Find the van't Hoff factor and degree of dissociation of the solute respectively.
Given :
$(K_{f})_{water} = 1.86 K k g m o l^{- 1} ρ_{w a t e r} = 1 g m L^{- 1}$