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

Dec 23, 2016

$\text{Mole fraction of KCl}$ $=$ $0.402$

#### Explanation:

$\text{Mole fraction of component}$ $=$ $\text{Moles of component"/"Moles of all components}$

$\text{Moles of KCl} = \frac{1.52 \cdot g}{74.55 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0204 \cdot m o l$

$\text{Moles of NaCl} = \frac{0.564 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1}$ $=$ $9.65 \times {10}^{-} 3 \cdot m o l$

$\text{Moles of LiCl} = \frac{0.857 \cdot g}{42.39 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0202 \cdot m o l$

$\text{Mole fraction of KCl} = \frac{0.0204 \cdot m o l}{\left(0.0202 + 0.00965 + 0.0204\right) \cdot m o l}$

$=$ $0.402$

Just as a check, the sum of the mole fractions of all the components should give unity.

$\text{Mole fraction of NaCl} = \frac{9.65 \times {10}^{-} 3 \cdot m o l}{\left(0.0202 + 0.00965 + 0.0204\right) \cdot m o l}$

$=$ $0.192$

$\text{Mole fraction of LiCl} = \frac{0.0202 \cdot m o l}{\left(0.0202 + 0.00965 + 0.0204\right) \cdot m o l}$

$=$ $0.402$

The sum of the mole fractions do give unity as required.