Question #2177b

Jul 18, 2017

${\chi}_{N a O H} \cong 0.05$

Explanation:

Now.....$\text{Molality}$ $=$ $\text{Moles of solute"/"Kilograms of solvent.}$

And our $N a O H \left(a q\right)$ solution is $3.0 \cdot \text{molal}$, i.e. $3.0 \cdot m o l \cdot k {g}^{-} 1$

Now ${\chi}_{N a O H} = \text{Moles of NaOH"/"Moles of NaOH + moles of water}$

$= \frac{3.0 \cdot m o l}{3.0 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}}$

$= \frac{3.0 \cdot m o l}{3.0 \cdot m o l + 55.52 \cdot m o l} = 0.0513$