Two beakers 1 and 2 containing 50g of 0.10m urea and 50g of 0.20M urea, respectively are placed under a tightly sealed bell jar at 298K. Calculate the mole fraction of urea in the solutions at equilibrium. Assume the ideal behavior

We require density of urea no. of moles of urea for solve this question 

If 50 gm take as 50mL and m take as M

Then no. of moles of urea

$=0.050\times 0.10+0.050\times 0.20$

= 0.005 + 0.01 = 0.015

No. of moles of water$=\frac{100}{18}=5.55$

${\mathrm{X}}_{\mathrm{urea}}=\frac{{\mathrm{n}}_{\mathrm{urea}}}{{\mathrm{n}}_{\mathrm{urea}}}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}$

$=\frac{0.015}{0.015+5.55}=\frac{0.015}{5.565}$

= 0.00269

= 0.00269

$=2.96\times {10}^{-3}$

For individuate of urea

${\mathrm{X}}_{\mathrm{urea}}=\frac{0.01}{0.01+5.55}=\frac{0.01}{5.65}=1.79\times {10}^{-3}$

$=1.8\times {10}^{-3}$