A 0-01 m aqueous solution of K 3- F e CN 6- was found to freeze at -0-062- C- If Kf for water is 1-86- C - molal then the apparent percentage of dissociation is-A- 78 -B- 100 -C- 50 -D- 60 -

The correct option is A 78 ∵ Δ T_f = iK_f m ∴ i= 0.062/0.01 × 1.86=0.062/0.0186=10/3 K_3[Fe(CN)_6] dissociates as, 3K^+ +[Fe(CN)_6]^3 ∴ i =(1 x)+3x+x/1=1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Motion Under Gravity

A 0.01 m aque...

Question

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:

60%

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

78%

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

100%

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

50%

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

Open in App

Solution

The correct option is B
78%

$∵ Δ T_{f} = i K_{f} m$

$∴ i = \frac{0.062}{0.01 \times 1.86} = \frac{0.062}{0.0186} = \frac{10}{3}$

$K_{3} [F e (C N)_{6}]$ dissociates as, $3 K^{+} + [F e (C N)_{6}]^{3 -}$

$∴ i = \frac{(1 - x) + 3 x + x}{1} = \frac{10}{3}$

$\Rightarrow$ 3 - 3x + 9x + 3x = 10 9x = 7

$\Rightarrow x = \frac{7}{9}$

$∴$ percentage dissociation $= \frac{700}{9} = 77.77 \approx 78 %$

Suggest Corrections

Similar questions

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]

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

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 freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is K_f = 1.86 k kg mol^-1)

Q. A

0.1

molal aqueous solution of a week acid is

30

% ionised. If

K_{f}

for water is

{1.86}^{\circ} C / m

, the freezing point of the solution will be:

Join BYJU'S Learning Program

A 0.01m aqueous solution of K3[Fe(CN)6] was found to freeze at –0.062∘C. If Kf for water is 1.86∘C /molal then the apparent percentage of dissociation is:

The correct option is B 78% ∵ ΔTf=iKfm ∴ i=0.0620.01×1.86=0.0620.0186=103 K3[Fe(CN)6] dissociates as, 3K++[Fe(CN)6]3− ∴ i=(1−x)+3x+x1=103 ⇒ 3 - 3x + 9x + 3x = 10 9x = 7 ⇒x=79 ∴ percentage dissociation =7009=77.77≈78%

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: