Question

An equimolar mixture of $\mathrm{NaOH}$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$ weighs $4$$\mathrm{g}$, then the weight of $\mathrm{NaOH}$ is:

Answer 1

Step 1: Suppose the weight of $\mathrm{NaOH}$ being ${\mathrm{w}}_1$​$\mathrm{g}$ and the weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3$being ${\mathrm{w}}_2$ ​$\mathrm{g}$.

The molar mass of $\mathrm{NaOH}$ is $40$$\mathrm{g}$$/\mathrm{mol}$

The molar mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ is $106$ $\mathrm{g}/\mathrm{mol}$

Step 2: Here, ${\mathrm{w}}_1+{\mathrm{w}}_2$$=$$4$

For an equimolar mixture,

$\frac{w_1}{40}=\frac{w_2}{106}$

$\Rightarrow$$\frac{w_1}{40}=\frac{4-w_1}{106}$

$\Rightarrow$$160-40w_1=106w_1$

$\Rightarrow$$160=106w_1+40w_1$

$\Rightarrow$$146w_1=160$

$\Rightarrow$$w_1=\frac{160}{146}$

$\Rightarrow$$w_1=1.096g$

Hence, the weight of $\mathrm{NaOH}$ is $1.096$ $\mathrm{g}$.