jee-main 2019 Q48

jee-main · India · session2_12apr_shift1 Not Maths
The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in $\text{mol kg}^{-1}$) of the aqueous solution is:
(1) $13.88 \times 10^{-3}$
(2) $13.88 \times 10^{-1}$
(3) $13.88$
(4) $13.88 \times 10^{-2}$
The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in $\text{mol kg}^{-1}$) of the aqueous solution is:\\
(1) $13.88 \times 10^{-3}$\\
(2) $13.88 \times 10^{-1}$\\
(3) $13.88$\\
(4) $13.88 \times 10^{-2}$