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Q.

A compound A (mol.wt = 600) forms two ions in solution. It contains 65% Pt, 23.6% Cl, 9.3% N and 2% H. Addition of silver nitrate to a solution of A does not give any precipitate immediately and no ammonia gas can be evolved from the solution. The number of chlorine atoms present in compound $A$ is (at. wts Pt – 195, Cl – 35.5, N – 14, H – 1)

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answer is 4.

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Detailed Solution

$\begin{array}{l} P t = \frac{65}{195} = \frac{1}{3} = 0.33 \\ C l = \frac{23.6}{35.5} = \frac{2}{3} = 0.67 \\ N = \frac{9.3}{14} = 0.66 \\ H = \frac{2}{1} = 2 \\ \frac{0.33}{0.33} : \frac{0.67}{0.33} : \frac{0.66}{0.33} : \frac{2}{0.33} \end{array}$
1: 2 : 2 : 6
Empirical $P t C l_{2} N_{2} H_{6}$
Empirical formula wt = 195+35.5+14+6 = 300
Molecular formula n × empirical formula $n = \frac{600}{300} = 2$
$= P t_{2} C l_{4} N_{4} H_{12}$

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