Neet Solved Paper 2013 Question 25
Question: An excess of $ AgN{O_3} $ is added to 100 mL of a 0.01 M solution of Dichlorotetraaquachromium $ \text{(III)} $ chloride. The number of moles of $ AgCl $ precipitate would be
Options:
A) 0.001
B) 0.002
C) 0.003
D) 0.01
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Answer:
Correct Answer: A
Solution:
The formula of dichlorotetraqua chromium (III), chloride is $ [Cr{{(H _2O)}_4}Cl _2]Cl. $
On ionisation it generates only one $ C{l^{-}} $ ion. $ \underset{\begin{smallmatrix} Initial \\ mmol \\ After,ionisation \end{smallmatrix}}{\mathop{[Cr{{(H _2O)}_4}}},\underset{\begin{smallmatrix} 100\times 0.01 \\ =1mmol \\ 0 \end{smallmatrix}}{\mathop{Cl _2]Cl}},\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} 0 \\ \\ 1mmol \end{smallmatrix}}{\mathop{[Cr{{(H _2O)}_4}}},Cl _2]\underset{\begin{smallmatrix} 0 \\ \\ 1mmol \end{smallmatrix}}{\mathop{^{+}+C{l^{-}}}}, $
One mole of $ C{l^{-}} $ ions react with only 1 mole of $ AgNO _3 $ molecule to produce 1 mole of $ AgCl $ .
$ \therefore $ 1 mmol or $ 1\times {10^{-3}} $ mole reacts with $ AgNO _3 $ to give $ AgCl $ $ =\frac{1\times 1\times {10^{-3}}}{1}={10^{-3}}or,0.001,mol,AgCl $
BETA
