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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 78gmol−1). The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

A
170gmol1
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B
270gmol1
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C
370gmol1
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D
140gmol1
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Solution

## 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

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