A) 0.2
B) 0.1
C) 0.02
D) 0.01
Correct Answer: C
Solution :
\[\text{Molarity}\,\text{=}\frac{\text{wt}\text{.}}{\text{mol}\text{.wt}\text{. }\!\!\times\!\!\text{ volume in litre}}.\] Mol. Wt. of oxalic acid \[\left( \underset{COOH}{\overset{COOH}{\mathop{|}}}\, \right).2{{H}_{2}}O\] \[=24+64+2+36=126\] Weight of oxalic acid = 12.6 Mol. wt.=126 Volume \[\text{=}\,\text{500}\,\text{mL}\,\text{= }\,\text{0}\text{.5 L}\] \[\text{Malrity}\,\text{=}\,\frac{12.6}{126\times 0.5}=\frac{1}{5}=0.2\] 0.2 M oxalic acid solution. Now, 10 mL of 0.2 M oxalic acid solution is diluted to 100mL solution, than from molarity equation. \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] \[0.2\times 10={{M}_{2}}\times 100\] \[{{M}_{2}}=\frac{0.2\times 10}{100}=0.02\]
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