1.

Dry air is passed through a solution containing 10 g of the solute in 90 g of water and then through pure water. The loss in weight of solution is 2.5 g and that of pure solvent is 0.05 g. Calculate the molecular weight of the solute.

Answer»

<P>50
180
100
25

SOLUTION :Loss in WEIGHT of solution `prop` V.P. of solution `(p_(s))`
Loss in weight of solvent `prop=p^(@)-p_(s)`
`THEREFORE""(p^(@)-p_(s))/(p_(s))=("Loss in weight of solvent")/("Loss in weight of solution")`
`=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2)M_(1))/(w_(1)M_(2))`
`=(0.05)/(2.5)=(10//M_(2))/(90//18)=(2)/(M_(2)) or M_(2)=100`


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