Molality of an aqueous solution of urea is \(4.44 \mathrm{~m}\). Mole fraction of urea in solution is \(x…

Molality of an aqueous solution of urea is \(4.44 \mathrm{~m}\). Mole fraction of urea in solution is \(x \times 10^{-3}\). Value of \(x\) is \(\qquad\) - (Integer answer)

Solution

Molality of urea is $4.44 \mathrm{~m}$, that means 4.44 moles of urea present in $1000 \mathrm{gm}$ of water. $\therefore \mathrm{X}_{\text {ura }}=\frac{4.44}{4.44+\frac{1000}{18}}$ $=0.0740$ OR $\begin{aligned} & 74 \times 10^{-3} \\ & X=74\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 2)

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