You visited us 24 times! Enjoying our articles? Unlock Full Access!

Depression in Freezing Point

0.1 M KI and ...

Question

$0.1 M K I$ and $0.2 M$ $A g N O_{3}$ are mixed in $3 : 1$ volume ratio. The depression of freezing point of the resulting solution will be [ $k_{f} (H_{2} O) = 1.86 K k g m o l^{- 1}$ ]:

3.72 K

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

1.86 K

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

0.93 K

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

0.279 K

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

Open in App

Solution

The correct option is C $0.279 K$
(d) $K I (a q) + A g N O_{3} (a q) + K N O_{3} (a q)$
$K I + A g N O_{3} \to K N O_{3} + A g I$
Initial : 0.3 mole 0.2 mole - -
Final : 0.1 mole - 0.2 0.2
Total moles of solute ions $= (0.1 + 0.2) \times 2 (∵ i = 2)$
[Solute] = $\frac{0.6}{4} M$
$Δ T_{f} = K_{f} \frac{0.6}{4}$
$= 1.86 \times \frac{0.6}{4} = 0.279 K$

Suggest Corrections

Similar questions

K_{f (H_{2} O)} = 1.8 K k g {m o l}^{- 1}

- {0.14}^{\circ} C

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

0.1 M H_{2} S O_{4} (a q)

- {0.3906}^{o} C

K_{2}

H_{2} S O_{4}

m = M

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

K_{f (H_{2} O)} = 1.86 K

^{- 1}

1 g

75 %

100 g

H_{2} O

{0.093}^{\circ} C

g

K_{f (H_{2} O)} = 1.86^{\circ} C / m

Join BYJU'S Learning Program