$12 g$ of urea is dissolved in $1 l i t r e$ of water and $68.4 g$ of sucrose is dissolved in $1 l i t r e$ of water. The lowering of vapour pressure of first case is

equal to second

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greater than second

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less than second

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double that of second

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Solution

The correct option is D equal to second
Moles of urea $= \frac{12}{60} = 0.2$

Moles of sucrose $= \frac{68.4}{342} = 0.2$

Both are non-electrolyte hence lowering of V.P. will be same.

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12g of urea is dissolved in 1 litre of water and 68.4 g of sucrose is dissolved in 1 litre of water. The lowering of vapour pressure of first case is

The correct option is D equal to secondMoles of urea =1260=0.2Moles of sucrose =68.4342=0.2Both are non-electrolyte hence lowering of V.P. will be same.

$12 g$ of urea is dissolved in $1 l i t r e$ of water and $68.4 g$ of sucrose is dissolved in $1 l i t r e$ of water. The lowering of vapour pressure of first case is

The correct option is D equal to second
Moles of urea $= \frac{12}{60} = 0.2$

Moles of sucrose $= \frac{68.4}{342} = 0.2$

Both are non-electrolyte hence lowering of V.P. will be same.