A) 1.2M
B) 1.8 M
C) 0.5 M
D) 0.4 M
Correct Answer: A
Solution :
2L of 3M \[{{C}_{6}}{{H}_{6}}\left( g \right)=0\]will contain 6 moles of \[{{H}_{2}}\left( g \right)={{p}_{2}}-3{{p}_{1}}\]. 3L of 1M \[{{C}_{6}}{{H}_{12}}\left( g \right)={{p}_{1}}\] will contain 3 moles of \[={{p}_{2}}-3{{p}_{1}}+{{p}_{1}}=30mm\] \[{{p}_{2}}-2{{p}_{1}}=30mm\] Thus, 6n moles of \[{{\text{p}}_{\text{1}}}\text{=10mm, }{{\text{p}}_{\text{2}}}\text{=50mm}\] will react with 3 moles of \[{{C}_{6}}{{H}_{6}}\] i.e. the two solution will react completely to from 3 moles of \[=\frac{10}{60}=\frac{1}{6}\] moles of \[=8\times \frac{1}{8}+6\times \frac{1}{2}=4\] ions in 2+3=5L solution Hence, molarity of \[=12\times \frac{1}{4}=3\] \[\text{C}{{\text{u}}_{\text{4}}}\text{A}{{\text{g}}_{\text{3}}}\text{Au}\text{.}\]
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