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  4. 12 g 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 :

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Q. $12 \,g$ 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 :

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Solution:

No. of moles of urea = mass $/$ moli mass $=12 / 60=0.2 \,mol$
No. of moles of sucrose $=68.4 / 342=0.2 \,mol$