A solution of sucrose (molar mass =342 ~g ~mol^ -1 ) has been prepared by dissolving 68.5 ~g of sucrose in 1000 ~g of water. The freezing point of the solution obtained will be (k_ f . for water =1.86 ~K ~kg ~mol^ -1 )

Correct answer: A. -0.372^ C Depression in freezing point, aligned & Δ T_ f =k_ f × m & where, aligned & m= molality = w_ B × 1000 M_ B · W_ A &=(68.5 × 1000)/(342 × 1000)=(68.5)/(342) & Δ T_ f =1.86 × (68.5)/(342) &= 0.372^ ^ C & Δ T_ f =T^0-T_s &= 0-0.372^ ^ C &=-0.372^ ^ C aligned aligned

NEET 2010

Exam: NEET
Year: 2010
Subject: Chemistry
Chapter: Solutions

NEET 2010 · Chemistry · Solutions

A solution of sucrose (molar mass =342 ~g ~mol^ -1 ) has been...

A solution of sucrose (molar mass

=342 \mathrm{~g} \mathrm{~mol}^{-1}

) has been prepared by dissolving

68.5 \mathrm{~g}

of sucrose in

1000 \mathrm{~g}

of water. The freezing point of the solution obtained will be \left(\mathrm{k}_{\mathrm{f}}\right.

for water

=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Options

A $-0.372^{\circ} \mathrm{C}$
B $-0.520^{\circ} \mathrm{C}$
C $+0.372^{\circ} \mathrm{C}$
D $-0.570^{\circ} \mathrm{C}$

Correct answer

A. $-0.372^{\circ} \mathrm{C}$

Step-by-step solution

Depression in freezing point,

\begin{aligned}& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{k}_{\mathrm{f}} \times \mathrm{m}\\ & \text{ where, } \begin{aligned}& \mathrm{m}=\text{ molality }=\frac{\mathrm{w}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \cdot \mathrm{W}_{\mathrm{A}}}\\ &=\frac{68.5 \times 1000}{342 \times 1000}=\frac{68.5}{342}\\ & \Delta \mathrm{T}_{\mathrm{f}}=1.86 \times \frac{68.5}{342}\\ &= 0.372^{{}^{\circ}} \mathrm{C}\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}^0-\mathrm{T}_s \\ &= 0-0.372^{{}^{\circ}} \mathrm{C}\\ &=-0.372^{{}^{\circ}} \mathrm{C} \end{aligned}

\end{aligned}$

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