`1g` of same non volatile solute is added to `100g` of two different solvents `A` and `B`. `K_(b)` of `A:B=1:5` Find out `((DeltaT_(b))_(A))/((DeltaT_(b))_(B))`
1g of a non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, (DeltaT_(b)(A))/(DeltaT_(b)(B)) is
Two solutions of non-volatile and non-electrolyte solute A and B are prepared separately. The molar mass ratio (M_(A))/(M_(B))=1/3 . Both are prepared as 5% by weight solution in water. Then what is the ratio of freezing point depresions, ((DeltaT_(f))_(A))/((DeltaT_(f))_(B)) of the solutions?
Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Variation of vapour pressure with temperaure and elevation in boiling point is marked. Ratio of DeltaT_(b)//K_(b) of 6% AB_(2) and 9% A_(2)B (AB_(2) and A_(2)B both are non-electrolytes) is 1 mol/kg in both cases. Hence, atomic masses of A and B are respectively.
when 1.80 g of a nonvolatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11K . If the boiling point of benzene is 353.23K and K_(b) for benzene is 2.53 KKg mol^(-1) , calculate the molecular mass of the solute. Strategy: From the boiling point of the solution, calculate the boiling point elevation, DeltaT_(b) , then solve the equation DeltaT_(b)=K_(b)m for the molality m . Molality equals moles of solute divided by kilograms of solvent (benzene). By substituting values for molality and kilograms C_(6)H_(6) , we can solve for moles of solute. The molar mass of solute equals mass of solute (1.80 g) divided by moles of solute. The molecular mass (in amu) has the same numerical value as molar mass in g mol^(-1) .
