Calculate the amount of water evaporated if $2.8 \ kg$ of $2 \ m$ $CaBr_2$ solution is heated to $110 \ ^oC$ at $1 \ atm$ .......... $gm$.
(Given: $K_b$ of water $= 0.5 \ K \ kg/mol$; Atomic mass of $Ca = 40$,$Br = 80$)

If an aqueous solution contains $9 \%$ and $1 \%$ $(w/w)$ of two non-volatile non-electrolytes $X$ (molecular weight $180$) and $Y$ (molecular weight $50$) respectively,the boiling point of the solution in ${ }^{\circ} C$ is approximately $(K_b = 0.52 \ K \ kg \ mol^{-1})$

When $3.3 \ g$ of an unknown substance is dissolved in $125 \ g$ of benzene (boiling point $80^\circ \text{C}$),the boiling point of the resulting solution is $80.66^\circ \text{C}$. What is the molar mass of the unknown substance? $(K_b = 3.28 \ \text{K kg mol}^{-1})$

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 $0.6\,K\,kg\,mol^{-1}$)

When $0.01 \ mol$ of nonvolatile solute is dissolved in a certain solvent,calculate the mass of the solvent in $kg$ if $\Delta T_{b} = 0.6 \ K$ and $K_{b}$ for the solvent $= 2 \ K \ kg \ mol^{-1}$. (in $kg$)

The solution containing $18 \ g \ dm^{-3}$ of glucose (molar mass $180 \ g \ mol^{-1}$) in water and another containing $6 \ g \ dm^{-3}$ of solute $A$ in water boil at the same temperature. What is the molar mass of $A$?