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Question

The weight of a non-volatile solute (m. wt. 40). Which should be dissolved in 114g octane to reduce its vapour pressure to 80%, is-

A
20 g
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B
30 g
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C
10 g
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D
40 g
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Solution

The correct option is C 10 g
According to Roults law of relative lowering of vapour pressure:
p01p1p01=w1/MWt1w1/M.wt1+w2/M.wt2
let relative vapour pressure of solvent octane p01 = x
vapour pressure of solute p1 = 80100x=0.8x
w1,M.wt1 of solute; w2,M.wt2 of solvent.
(x0.8x)x=w1/40w1/40+(114/114)
0.8w140=0.2
the weight of solute required = 10g


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