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Standard XII

Chemistry

Van't Hoff Factor

A 0.15 mola...

Question

A $0.15$ molal solution of $K_{4} [F e (C N)_{6}]$ in water freezes at $- {0.65}^{\circ} C$ . What is the apparent percentage of dissociation of this compound in this solution?
$[K_{f}$ for water $= {1.86}^{\circ} C m o l^{- 1}]$

0.33

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0.52

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0.63

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0.79

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Solution

The correct option is A $0.33$
We know,
$Δ T_{f} = i K_{f} m$

$0.65 = i \times 1.86 \times 0.15$

$K_{4} [F e (C N)_{6}] ⇋ 4 K^{+} + [F e (C N)_{6}]^{4 -}$

$i = 2.33 = 1 + 4 α$

$α = 0.33$ .

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Similar questions

A 0.01m aqueous solution of $K_{3} [F e (C N)_{6}]$ was found to freeze at $- {0.062}^{\circ} C$ . If Kf for water is ${1.86}^{\circ} C$ /molal then the apparent percentage of dissociation is:

Q. A

0.01 m

aqueous solution of

K_{3} [F e (C N)_{6}]

freezes at

- {0.062}^{\circ} C

. What is the apparent percentage of dissociation?

[K_{f}

for water

= 1.8]

[Note : Write the value to the nearest integer]

K_{f}

for water is

{1.86}^{\circ} C k g m o l^{- 1} .

What is the molality of a solution which freezes at

- {0.192}^{\circ} C

Q. 0.01 m aqueous solution of

K_{3} [F e {(C N)}_{6}]

freezes at

- 0.0 62^{\circ} C

, what is the apparent percentage of dissociation ?

(k_{f} (H_{2} O) = 1.86 K k g {m o I}^{- 1})

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

- {0.30}^{o} 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. The vapour pressure of this solution at 298 K in cm of Hg to the nearest integer is________.

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A 0.15 molal solution of K4[Fe(CN)6] in water freezes at −0.65∘C. What is the apparent percentage of dissociation of this compound in this solution? [Kf for water =1.86∘C mol−1]

The correct option is A 0.33We know,ΔTf=iKfm0.65=i×1.86×0.15K4[Fe(CN)6]⇋4K++[Fe(CN)6]4−i=2.33=1+4αα=0.33.

A $0.15$ molal solution of $K_{4} [F e (C N)_{6}]$ in water freezes at $- {0.65}^{\circ} C$ . What is the apparent percentage of dissociation of this compound in this solution?
$[K_{f}$ for water $= {1.86}^{\circ} C m o l^{- 1}]$

The correct option is A $0.33$
We know,
$Δ T_{f} = i K_{f} m$

$0.65 = i \times 1.86 \times 0.15$

$K_{4} [F e (C N)_{6}] ⇋ 4 K^{+} + [F e (C N)_{6}]^{4 -}$

$i = 2.33 = 1 + 4 α$

$α = 0.33$ .