A solution containing 0.2563 g of naphthalene (molecular mass = 128) in 50 g of

Question

A solution containing 0.2563 g of naphthalene (molecular mass = 128) in 50 g of carbon tetrachloride yields a boiling point elevation of

{0.201}^{\circ} C

while a solution of 0.6216 g of an unknown solute in the same mass of the solvent gives a boiling point elevation of

{0.647}^{\circ} C

. Find the molecular mass of unknown solute.

Options:

A . 96.44

B . 86.44

C . 90

D . 88

Answer: Option A
:
A
We know that

K_{b} = \frac{Δ T_{b} \times W_{1} \times M w_{2}}{1000 \times W_{2}}

First let’s calculate value of

K_{b} f o r C C l_{4}

.
For

C C l_{4};

K_{b} = \frac{0.201 \times 50 \times 128}{1000 \times 0.2563} = 5.019 k k g m o l^{- 1}

Now, for calculating mass of unknown solute, we can use the same value of

K_{b}

M w_{2} = \frac{1000 \times K_{b} \times W_{2}}{Δ T_{b} \times W_{1}}

\frac{1000 \times 5.019 \times 0.6216}{0.647 \times 50}

= 96.44

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