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 to $\mathrm{NaNO}_3$ 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 $x$ gives,
$\begin{aligned} & x=3.3 \mathrm{~g} \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \\ & \Rightarrow \quad \mathrm{NaNO}_3=1.7 \mathrm{~g} \text {. } \\ & \end{aligned}$
BETA
