Lowering in vapour pressure is highest for:

0.2 m

urea

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0.1 m

glucose

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0.1 m M g S O_{4}

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0.1 m B a C l_{2}

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Solution

The correct option is A $0.2 m$ urea
Lowering in vapour pressure is directly proportional
to molality which in turn is directly proportional
to n factor (no. of ions it gives)
The concentration of urea, i.e, $0.2 M$ is much
more than glucose, $M g S O_{4}$ & $B a C l_{2}$
Therefore urea will have greater molality &
therefore it will have highest vapour pressure.
Ans is A) $0.2 M$ urea

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