Some Basic Concepts of Chemistry 1 Question 31
51. A solid mixture $(5.0 \mathrm{~g})$ consisting of lead nitrate and sodium nitrate was heated below $600^{\circ} \mathrm{C}$ until the weight of the residue was constant. If the loss in weight is 28.0 per cent, find the amount of lead nitrate and sodium nitrate in the mixture.
$(1990,4 \mathrm{M})$
Show Answer
Solution:
- Heating below $600^{\circ} \mathrm{C}$ converts $\mathrm{Pb}\left(\mathrm{NO}{3}\right){2}$ into $\mathrm{PbO}$ but to $\mathrm{NaNO}{3}$ into $\mathrm{NaNO}{2}$ as
MW
$$ \begin{array}{cl} \mathrm{Pb}\left(\mathrm{NO}{3}\right){2} & \stackrel{\Delta}{\longrightarrow} \mathrm{PbO}(s)+2 \mathrm{NO}{2} \uparrow+\frac{1}{2} \mathrm{O}{2} \uparrow \ 330 & 222 \ \mathrm{NaNO}{3} & \stackrel{\Delta}{\longrightarrow} \mathrm{NaNO}{2}(s)+\frac{1}{2} \mathrm{O}_{2} \uparrow \end{array} $$
MW : $\quad 330$
MW : $\quad 85 \quad 28 \quad 69$
Weight loss $=5 \times \frac{28}{100}=1.4 \mathrm{~g}$
$\Rightarrow$ Weight of residue left $=5-1.4=3.6 \mathrm{~g}$
Now, let the original mixture contain $x \mathrm{~g}$ of $\mathrm{Pb}\left(\mathrm{NO}{3}\right){2}$.
$\because \quad 330 \mathrm{~g} \mathrm{~Pb}\left(\mathrm{NO}{3}\right){2}$ gives $222 \mathrm{~g}$ PbO
$\therefore \quad x \mathrm{~g} \mathrm{~Pb}\left(\mathrm{NO}{3}\right){2}$ will give $\frac{222 x}{330} \mathrm{~g} \mathrm{PbO}$
Similarly, $85 \mathrm{~g} \mathrm{NaNO}{3}$ gives $69 \mathrm{~g} \mathrm{NaNO}{ }{2}$
$\Rightarrow \quad(5-x) \mathrm{g} \mathrm{NaNO}{3}$ will give $\frac{69(5-x)}{85} \mathrm{~g} \mathrm{NaNO}{2}$
$\Rightarrow$ Residue : $\frac{222 x}{330}+\frac{69(5-x)}{85}=3.6 \mathrm{~g}$
Solving for $x$ gives, $\quad x=3.3 \mathrm{~g} \mathrm{~Pb}\left(\mathrm{NO}{3}\right){2}$
$\Rightarrow \quad \mathrm{NaNO}_{3}=1.7 \mathrm{~g}$.
