Question

If 10% (w/w) aqueous solution of $K_{2}C{O}_3$ is kept in a container whose volume capacity is 100 L then, mole fraction ($x_{K_{2}C{O}_3}$) is equal to:

• A. 0.0188
• B. 0.0368
• C. 0.0254
• D. 0.0138

Answer 1

The correct option is D 0.0138

10 %(w/w) means 10 g of $K_{2}C{O}_3$ is present in 100 g of solution.
Mass of $K_{2}C{O}_3$ in 100 g solution = 10 g
Number of moles of $K_{2}C{O}_3$ in 100 g solution $=\frac{givenmass}{molarmass}=\frac{10}{138}=0.07mol$
Weight of solvent (water) in 100 g solution $=100-10=90g$
Number of moles of water in 100 g solution $=\frac{90}{18}=5mol$
$Molefraction=\frac{molesof{K}_{2}C{O}_3}{molesof{K}_{2}C{O}_3+molesofsolvent}$
$Molefractionof{K}_{2}C{O}_3=\frac{0.07}{(0.07+5)}=0.0138$