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Question

# 12 g of non volatile solute having (Molar mass=120 g/mol) was dissolved in 9.9 mol of solvent. If the vapour pressure of pure solvent is P∘A, the vapour pressure of solution is:

A
0.99 PA
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B
0.01 PA
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C
0.12 PA
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D
0.88 PA
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Solution

## The correct option is A 0.99 P∘AWhen a non volatile solute is added to a volatile solvent, vapour pressure is lowered. By relative lowering of vapour pressure formula: P∘A−PSP∘A=xsolute⇒P∘A−PSP∘A=nAnA+nB nA are the number of moles of solute=12120=0.1 mol nB are the number of moles of solvent=9.9 mol ⇒P∘A−PSP∘A=0.10.1+9.9⇒P∘A−PSP∘A=0.01⇒PS=0.99 P∘A

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