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........

A molecule A_(x) dissovle in water and is non volatile. A solution of

1.	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» Solution :`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`