You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.52

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.63

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.79

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

$0.65 = i \times 1.86 \times 0.15$

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

$i = 2.33 = 1 + 4 α$

$α = 0.33$ .

Suggest Corrections

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________.

Join BYJU'S Learning Program

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,
$Δ T_{f} = i K_{f} m$

$0.65 = i \times 1.86 \times 0.15$

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

$i = 2.33 = 1 + 4 α$

$α = 0.33$ .