`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

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?

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?

Consider following figure and answer the questions at the end of it. Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Variation of vapour pressure with temperature and showing elevation in boiling point. Given that DeltaT_(b) is the elevation in boiling point of the solvent in a solution of molality m then underset(mto0)(Lt)((DeltaT_(b))/(m)) is equal to