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Question

The mass of a non-volatile solute of molar mass 40 g mol1 that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is :

A
10 g
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B
11.4 g
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C
9.8 g
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D
12.8 g
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Solution

The correct option is C 10 g
The mass of a non-volatile solute of molar mass 40 g /mol that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is 10 g.
Let w g of the non-volatile solute is used. Its molar mass is 40 g/mol.

The number of moles of non volatile solute =w40=0.025w moles.

114 g of octane (molar mass 114 g/mol) corresponds to 114114=1 mole.

The mole fraction of non volatile solute is =0.025w1+0.025w. The mole fraction of non volatile solute is proportional to the relative lowering in the vapour pressure.

P0PP0=XB

20100=0.025w1+0.025 w

100(0.025w)=20(1+0.025w)

2.5w=20+0.50w

2w=20

w=10 g

Hence, the correct option is A

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