Question

Calculate the molality and mole fraction of components in $10\%$ w/w aqueous sodium hydroxide solution.

Answer 1

$10\%w/w$ $NaOH$

$10gNaOH$ and $90g{H}_{2}O$

$\Rightarrow M=\frac{wt}{mol.wt}\times \frac{1000}{wtofsolvent\left(g\right)}$

$=\frac{10}{40}\times \frac{1000}{90}$

$=2.78mol/kg$

$\Rightarrow x=\frac{n_{NaOH}}{n_{NaOH}+n_{H_{2}O}}=\frac{\left(\frac{10}{40}\right)}{\left(\frac{10}{40}\right)+\left(\frac{90}{18}\right)}$

$=\frac{0.25}{0.25+5}$

$=0.0476$

Hence, the answer is $\mathrm{0.0476.}$