1.

Dry air was passed successively through a solution of 5 gm of solute dissolved in 80.0 gm of water and through pure water. The loss in weight of the solution was 2.5 gm and that of the pure solvent was 0.04 gm. What is the molecular weight of the solute?

Answer»

SOLUTION :`p ALPHA 2.5 GM,(p^(@)-p)alpha0.04 gm`
`{:("The relative lowring of"),("vapour pressure"):}}=(p^(@)-p)/(p^(@))`
`=(W_(2))/(M_(2)).(M_(1))/(W_(1))`
`THEREFORE (0.04)/(2.54)=(5xx18)/(M_(2)xx80)(thereforeM_(2)=71.43)`
`M_(2) =" mol.wt of solute "= 71.43`


Discussion

No Comment Found