$0.15 \ g$ of a solute is dissolved in $15 \ g$ of a solvent. If the solution boils at a temperature $0.215 \ K$ higher than the boiling point of the pure solvent,what is the molar mass of the solute? $(K_b = 2.15 \ K \ kg \ mol^{-1})$

Calculate the molar mass of a non-volatile solute when $5 \ g$ of it is dissolved in $50 \ g$ of solvent,which boils at $119.6^{\circ} C$. $[K_{b} = 3.2 \ K \ kg \ mol^{-1}$,boiling point of pure solvent $= 118^{\circ} C]$.

$2 \, g$ of a non-volatile non-electrolyte solute is dissolved in $200 \, g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1: 8$. The elevation in boiling points of $A$ and $B$ are in the ratio $\frac{x}{y} (x: y)$. The value of $y$ is .... (Nearest integer)

$A$ solution of non-volatile solute has a boiling point elevation of $1.75 \ K$. Calculate the molality of the solution $[K_{b} = 3.5 \ K \ kg \ mol^{-1}]$.

$11.1 \ g$ of $CaCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. $[K_b = 0.5 \ K \ kg \ mol^{-1}]$ :-

$1 \ g$ 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