Some Basic Concepts of Chemistry - Result Question 52
51. A solid mixture $(5.0 g)$ consisting of lead nitrate and sodium nitrate was heated below $600^{\circ} C$ 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 M)
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Answer:
Correct Answer: 51. (1.7 g)
Solution:
Heating below $600^{\circ} \mathrm{C}$ converts $\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$ into PbO but $\mathrm{NaNO}_3$ into $\mathrm{NaNO}_2$ as
$\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \xrightarrow{\Delta} \mathrm{PbO}(s)+2 \mathrm{NO}_2 \uparrow+\frac{1}{2} \mathrm{O}_2 \uparrow$
$\Rightarrow$ 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 g PbO
$\therefore x \mathrm{~g} \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2$ 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 $\frac{69(5-x)}{85} \mathrm{~g} \mathrm{NaNO}_2$
$\Rightarrow \quad \text { Residue : } \frac{222 x}{330}+\frac{69(5-x)}{85}=3.6 \mathrm{~g}$
Solving for $x$, we get,
$\begin{aligned} & x=3.3 \mathrm{~g} \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \\ & \Rightarrow \quad \mathrm{NaNO}_3=1.7 \mathrm{~g} \text {. } \\ & \end{aligned}$
BETA
