The weight in grams of a non-volatile solute (mol.wt.60) to be dissolved in 90 g of water to produce a relative lowering of vapour pressure of 0.02 is

Options

A. 4
B. 8
C. 6
D. 10

Answer

( c)

Explanation

`(P_(o)-P_(s))/(P_(o))= (w_(2)/m_(2))/(w_(1)/(m_(1))+(w_(2))/m_(2))`
`0.02=(w_(2)/60)/(90/(18)+(w_(2))/(60))`
`(1)/(0.02)=(w_(2)/60)/(90/02)=(90/(18)+(w_(2))/(60))/(w_(2)/(6)) =(5+(w_(2))/(60))/(w_(2)/(60))=(300+w_(2))/(w_(2))` `(1)/(0.02)=(w_(2)/60)/(90/02)=(90/(18)+(w_(2))/(60))/(w_(2)/(6)) =(5+(w_(2))/(60))/(w_(2)/(60))=(300+w_(2))/(w_(2))`