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-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-C- Find the molecular mass of unknown solute-

We know that K_b=Δ T_b × W_1 × Mw_2/1000 × W_2 First let’s calculate value of K_b for CCl_4. For CCl_4; K_b=0.201 × 50 × 128/1000 × 0.2563=5.019 k kg mol^ 1 Now ...

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Byju's Answer

Standard XII

Chemistry

Elevation in Boiling Point

A solution co...

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.

96.44

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86.44

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Solution

The correct option is A 96.44
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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