$0.1 \ molal$ aqueous solution of glucose boils at $100.16^{\circ} C$. What is the boiling point of a $0.5 \ molal$ aqueous solution of glucose (in $^{\circ} C$)?

The boiling point of water $(100\,^{\circ}C)$ becomes $100.52\,^{\circ}C$,if $3\,g$ of a non-volatile solute is dissolved in $200\,mL$ of water. The molecular weight of the solute is: (${K_b}$ for water is $0.6\,K\,kg\,mol^{-1}$)

Boiling point of a $2\%$ aqueous solution of a non-volatile solute $A$ is equal to the boiling point of an $8\%$ aqueous solution of a non-volatile solute $B$. The relation between molecular weights of $A$ and $B$ is:

$1 \ g$ of non-volatile non-electrolyte solute is dissolved in $100 \ g$ 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,$\frac{\Delta T_b (A)}{\Delta T_b (B)}$ is

An aqueous solution containing $1 \ g$ of urea boils at $100.25 \ ^oC$. The aqueous solution containing $3 \ g$ of glucose in the same volume will boil at ........ $^oC$. (Molecular weight of urea and glucose are $60$ and $180$ respectively)

$x \ g$ of urea (molar mass $60 \ g \ mol^{-1}$) is completely dissolved in $y \ g$ of pure water and the solution boiled at $373.202 \ K$. If the boiling point of pure water at $1.013 \ bar$ is $373.15 \ K$, then $x:y$ is $(K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1})$