A molecule `A_(x)` dissovle in water and is non volatile. A solution of certain molality showed a depression of `0.93K` in freezing point. The same solution boiled at `100.26^(@)C`. When `7.87g` of `A_(x)` was dissovled in `100g` of water, the solution boiled ate `100.44^(@)C`. Given `K_(f)` for water `=1.86K kg "mol"^(-1)` atomic mas of `A=31u`. Assume no association or dissociatioin of solute. Calculate the value of `x`........

Answer

3

Explanation

`DeltaT_(f)=K_(f)m`
`m=0.93/1.86=0.5`
`DeltaT_(b)=K_(b)m`
`K_(b)=0.26/0.5=0.52K kg "mol"^(-1)`
Similarly `DeltaT_(b)^(')=K_(b)m^(')`
`DeltaT_(b)^(')=(K_(b)xxwxx1000)/(Wxxm.wt)`
`:.` m.wt of solute `=(K_(b)xxwxx1000)/(wxxDelta_(b)^('))`
`=(0.52xx7.87xx1000)/(100xx0.44)`
`31x~~93`
`x=3`