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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

The vapour pr...

Question

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass $78 g m o l^{- 1})$ . The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

170 g m o l^{- 1}

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270 g m o l^{- 1}

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370 g m o l^{- 1}

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140 g m o l^{- 1}

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Solution

The correct option is A $170 g m o l^{- 1}$
The given values are $P^{\circ} = 0.850 b a r, P s = 0.845 b a r; M w_{1} = 78 g m o l^{- 1}$ $W_{2} = 0.5 g; W_{1} = 39 g$ Now using equation
$\frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{W_{1} \times M w_{1}}{M w_{2} \times W_{1}}$ On substituting all the given values, we get $\frac{0.850 - 0.845}{0.850} = \frac{0.5 g \times 78 g m o l^{- 1}}{M w_{2} \times 39 g}$ $∴ M w_{2} = 170 g m o l^{- 1}$

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The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78gmol−1). The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

The correct option is A 170gmol−1The given values are P∘=0.850bar,Ps=0.845bar;Mw1=78gmol−1 W2=0.5g;W1=39g Now using equation P∘−PsP∘=W1×Mw1Mw2×W1 On substituting all the given values, we get 0.850−0.8450.850=0.5g×78gmol−1Mw2×39g ∴Mw2=170gmol−1

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass $78 g m o l^{- 1})$ . The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?