A) 4mL
B) 8 mL
C) 40 mL
D) 80 mL
Correct Answer: C
Solution :
Given, density of \[C{{H}_{3}}OH=0.8\,kg\,{{L}^{-1}}\] Molarity\[=0.25\,M\] . Volume of \[0.25\,M=4L\] Volume needed\[~=?\] First of all we find mass of methanol (i.e. given mass) \[\text{Molarity}\,\text{=}\frac{\text{Number}\,\text{of}\,\text{moles}}{\text{Volumeof}\,\text{solution(L)}}\] \[\text{Molarity}\,\,\text{=}\,\,\frac{\text{Given}\,\text{mass}}{\text{Molar}\,\text{mass}}\,\,\text{ }\!\!\times\!\!\text{ }\,\frac{\text{1}}{\text{Volume}\,\text{of}\,\text{solution(L)}}\]Molar mass of \[C{{H}_{3}}OH=12+3+16+1=32\,\text{mo}{{\text{l}}^{-1}}\] \[\therefore \] \[0.25\,mo{{l}^{-1}}=\frac{\text{Given}\,\text{mass}}{32\,gmo{{l}^{-1}}}\times \,\frac{1}{4L}\] \[\therefore \]Given mass \[=32\,g\]or \[0.032\,kg\] Again \[\because \] \[denstiy=\frac{given\,mass\,(kg)}{volume\,(mL)}\] \[\therefore \] \[0.8\,kg\,{{L}^{-1}}=\frac{0.032\,kg}{V(mL)}\] or \[0.8\times 1000\,kg\,m{{L}^{-1}}=\frac{0.032\,kg}{V(mL)}\] \[\therefore \] \[V(mL)=\frac{0.032\,kg}{0.8\,kg\,{{L}^{-1}}}\,\times \,1000\] \[V(mL)=40\]
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