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Real Gases

On dissolving...

Question

On dissolving $3.24 g$ of sulphur in $40 g$ of benzene, the boiling point of the solution was higher than that of benzene by $0.81 K$ . What is the molecular formula of sulphur?
(Given, $K_{b}$ for benzene $= 2.53 K k g m o l^{- 1}$ , atomic mass of sulphur $= 32 g m o l^{- 1}$ )

S_{4}

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S_{8}

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S_{2}

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S_{6}

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Solution

The correct option is B $S_{8}$
The given values are:
weight of solute, $W_{B} = 3.24 g$
weight of solvent, $W_{A} = 40 g$
$△ T_{b} = 0.81 K; K_{b} = 2.53 K k g m o l^{- 1}$
Using the formula,
$M_{B} = \frac{K_{b} \times 1000 \times W_{B}}{△ T_{b} \times W_{A}}$
we get,
$∴ M_{B} = \frac{2.53 \times 1000 \times 3.24}{0.81 \times 40} = 253$

Let the molecular formula of sulphur be $S_{x}$
The atomic mass of sulphur $= 32$
Molecular mass $= 32 \times x$
$∴ 32 x = 253$
$x = \frac{253}{32} = 7.91 \approx 8$
So, the molecular formula of sulphur $= S_{8}$

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On dissolving 3.24 g of sulphur in 40 g of benzene, the boiling point of the solution was higher than that of benzene by 0.81 K. What is the molecular formula of sulphur? (Given, Kb for benzene =2.53 Kkgmol−1, atomic mass of sulphur =32 gmol−1)

On dissolving $3.24 g$ of sulphur in $40 g$ of benzene, the boiling point of the solution was higher than that of benzene by $0.81 K$ . What is the molecular formula of sulphur?
(Given, $K_{b}$ for benzene $= 2.53 K k g m o l^{- 1}$ , atomic mass of sulphur $= 32 g m o l^{- 1}$ )