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Depression in Freezing Point

A 0.1 molal a...

Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If $K_{f}$ for water is ${1.86}^{o} C / m$ , the freezing point of the solution will be:

- {0.24}^{o} C

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- {0.18}^{o} C

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- {0.54}^{o} C

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- {0.36}^{o} C

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Solution

The correct option is A $- {0.24}^{o} C$
$Δ T_{f} = i * m * K_{f} w h e r e, i = V a n t H o f f^{'} s F a c t o r K_{f} = F r e e z i n g p o i n t d e p r e s s i o n c o n s t a n t m = M o l a l i t y o f s o l u t i o n . i = 1 + (n - 1) * α w h e r e α = D e g r e e o f d i s s o c i a t i o n . S o, i = 1 + (2 - 1) * 0.3 = 1.3 {Δ T}_{f} = 1.3 * 1.86 * 0.1 = 0.24 T_{f} = T^{0} - {Δ T}_{f} = 0 - 0.24 = - {0.24}^{0} C$

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Similar questions

Q. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If

K_{f}

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1.86^{o} C / m

, the freezing point of the solution will be:

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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86oC/m, the freezing point of the solution will be:

The correct option is A −0.24oCΔTf=i∗m∗Kfwhere,i=VantHoff′sFactorKf=Freezingpointdepressionconstantm=Molalityofsolution.i=1+(n−1)∗αwhereα=Degreeofdissociation.So,i=1+(2−1)∗0.3=1.3ΔTf=1.3∗1.86∗0.1=0.24Tf=T0−ΔTf=0−0.24=−0.240C

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If $K_{f}$ for water is ${1.86}^{o} C / m$ , the freezing point of the solution will be: