1.

Dry air was passed through a solution containing 20g of a substance in 100-0g of water and then through pure water. The loss in mass of the solution was 2.945g and that of pure water was 0.059g. Calculate the molar mass of the substance.

Answer»

SOLUTION :We have,
`("loss in mass of PURE water")/("loss in mass of solution")=(p^(0)-p)/(p)`
or `(p^(0)-p)/(p)=(0.059)/(2.945)=0.02`
or `(p^(0))/(p)-1=0.02`
or `(p^(0))/(p)=1.02`
`(p)/(p^(0))=(1)/(1.02)`
`(p^(0)-p)/(p^(0))=1-(1)/(1.02)=0.0196`
`:.(N)/(N)=(20//M)/(100//18)=0.0196`, `M=183.6g//"MOLES"`


Discussion

No Comment Found

Related InterviewSolutions