# What is the mole fraction of LiCI in a mixture of 0.564g NaCl, 1.52g KCl and 0.857g LiCI?

Jul 11, 2017

Mole fraction $\left({X}_{c o m p o u n d}\right)$ = $\left(\text{moles compound"/"total moles of all compounds}\right)$

Given:
$0.564 g N a C l$ = $\frac{0.564 g}{58 \left(\frac{g}{\text{mol}}\right)}$ = $0.0097 \text{mole}$

$1.52 g K C l$ = $\frac{1.52 g}{74 \left(\frac{g}{\text{mol}}\right)}$ = $0.0205 \text{mole}$

$0.857 g L i C l$ = $\frac{0.857 g}{42 \left(\frac{g}{\text{mol}}\right)}$ = $0.0204 \text{mole}$

$\Sigma \text{moles" = (0.0097 + 0.0205 + 0.0204)"mole}$ = $\text{0.0506mole}$

${X}_{L i C l}$ = $\left(\text{moles LiCl")/(Sigma"moles}\right)$ = $\frac{0.0204}{0.0506}$ = $0.04032$