An aqueous solution containing a non-volatile solute boils at 101- C- What is the freezing point of the same solution-GivenKf-1-86- C-m and Kb-0-51- C-m

Δ T_b=K_b× m Here m is molality m=Δ T_bK_b= 101 1000.51=10.51 Also, Δ T_f=K_f× m m=Δ T_fK_f Put the value of m ⇒10.51=Δ T_f1.86 Δ T_f=3.647^∘ C T_f^0 T_f=3.647 ...

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Standard XII

Chemistry

Depression in Freezing Point

An aqueous so...

Question

An aqueous solution containing a non-volatile solute boils at $101^{\circ} C .$ What is the freezing point of the same solution?
Given
$(K_{f} = {1.86}^{\circ} C / m$ and $K_{b} = {0.51}^{\circ} C / m)$

{3.647}^{\circ} C

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- {3.647}^{\circ} C

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- {0.364}^{\circ} C

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+ {0.364}^{\circ} C

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Solution

The correct option is B $- {3.647}^{\circ} C$
$Δ T_{b} = K_{b} \times m$
Here m is molality
$m = \frac{Δ T_{b}}{K_{b}} = \frac{101 - 100}{0.51} = \frac{1}{0.51}$
Also,
$Δ T_{f} = K_{f} \times m$
$m = \frac{Δ T_{f}}{K_{f}}$
Put the value of m
$\Rightarrow \frac{1}{0.51} = \frac{Δ T_{f}}{1.86}$

$Δ T_{f} = {3.647}^{\circ} C T_{f}^{0} - T_{f} = {3.647}^{\circ} C 0 - T_{f} = {3.647}^{\circ} C T_{f} = - {3.647}^{\circ} C$

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Depression in Freezing Point

Standard XII Chemistry

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An aqueous solution containing a non-volatile solute boils at 101∘C. What is the freezing point of the same solution? Given (Kf=1.86∘C/m and Kb=0.51∘C/m)

The correct option is B −3.647∘CΔTb=Kb×m Here m is molality m=ΔTbKb=101−1000.51=10.51 Also, ΔTf=Kf×m m=ΔTfKf Put the value of m ⇒10.51=ΔTf1.86 ΔTf=3.647∘CT0f−Tf=3.647∘C0−Tf=3.647∘CTf=−3.647∘C

An aqueous solution containing a non-volatile solute boils at $101^{\circ} C .$ What is the freezing point of the same solution?
Given
$(K_{f} = {1.86}^{\circ} C / m$ and $K_{b} = {0.51}^{\circ} C / m)$