A) 1.15 cc
B) 2 cc
C) 2.15 cc
D) 2.30 cc
Correct Answer: B
Solution :
Molarity \[\text{=}\,\,\frac{\text{mass }\!\!\times\!\!\text{ 100}}{\text{molecular}\,\text{weight}\,\,\text{ }\!\!\times\!\!\text{ }\,\text{volume}\,\text{of}\,\text{solution}}\] Molecular weight of ethyl alcohol, \[{{C}_{2}}{{H}_{5}}OH=12\times 2+5+16+1\] \[=24+22=46\,\text{g}\,\text{mo}{{\text{l}}^{-1}}\] On substituting the values, we get \[0.5\,M=\frac{mass\times 1000}{46\times 100}\] \[Mass=\frac{0.5\times 46}{10}=2.3\,g\] \[\therefore \] Volume of ethyl alcohol required, \[\text{V}\,\text{=}\frac{\text{mass}\,\text{of}\,\text{ethyl}\,\text{alcohol}}{\text{density}\,\text{of}\,\text{ethyl}\,\text{alcohol}}=\frac{2.3\,g}{1.15\,g/cc}=2cc\]
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