If $10 \ g$ of a solute (molar mass $= 100 \ g/mol$) is dissolved in $100 \ g$ of water,then the molal elevation constant $K_b$ of the solvent will be equal to:

$2 \text{ g}$ of a non-electrolyte solute, when dissolved in $50 \text{ g}$ of water, increases its boiling point by $0.13 \text{ K}$. Calculate the molar mass of the solute if the molal elevation constant $(K_b)$ of water is $0.52 \text{ K kg mol}^{-1}$.

$1 \ g$ 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

The boiling point of a $0.1 \ m$ aqueous solution of urea is $100.18 \ ^\circ C$ at $1 \ atm$. The molal elevation constant $(K_b)$ of water is:

Which one of the following would produce maximum elevation in boiling point?

$A$ solution containing $12.5 \ g$ of non-electrolyte substance in $250 \ g$ of water shows boiling point elevation of $0.52 \ K$. Calculate the molar mass of the substance. $(K_b = 0.52 \ K \ kg \ mol^{-1})$ ......... $g \ mol^{-1}$.