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The amount of solute (mol. Mass=60) that...

The amount of solute (mol. Mass=60) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is

A

30 g

B

60 g

C

120 g

D

12 g

Text Solution

Verified by Experts

The correct Answer is:
B
Relative lowering of vapour pressure is given by the formula
`(p^(o)-p_(s))/(p^(o)) = (w_(A))/(M_(A)) xx (M_(B))/(w_(B))`
As vapour pressure of water is lowered by `10%`.
`therefore (p^(o) - p_(s))/(p^(o)) = (10)/(100)`
`therefore (10)/(100) = (w_(A))/(60) xx (18)/(180)`
or `w_(A) = 60 g`
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