A piece of plumber's solder weighing 3.0gm was dissolved in dilute nitric acid, then treated with dilute $H_{2} {SO}_{4}$ . This precipitated the lead as ${PbSO}_{4}$ , which after washing and drying weighed 2.93gm.The solution was then neutralized to precipitate stannic acid, which was decomposed by heating, yielding $1.27 gm {SnO}_{2}$ . What is the analysis of the solder? $Pb = 207.2 amu, Sn = 118.7 amu)$ .

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$66.7 % Pb, 33.3 % Sn$

$33.3 % Pb, 66.7 % Sn$

$1 gmPb, 2 gmSn$

$2 gmPb, 1 gmSn$

answer is A, D.

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Detailed Solution

Percentage of lead

$\begin{array}{l} % Pb = \frac{2.93}{303.2} \times \frac{1}{3} \times 100 \times 207.2 \\ = 66.7 %; % S_{n} = 33.3 % \\ = 2 g = 1 g \end{array}$

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