A) 4 mL
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 =4 L Volume needed = ? First of all we find mass of methanol (i.e. given mass) \[Molarity=\frac{Number\text{ }of\text{ }moles}{Volume\text{ }of\text{ }solution\,(L)}\] \[Molarity=\frac{Given\text{ }mass}{Molar\text{ }mass}\times \frac{1}{Volume\text{ }of\text{ }solution\,(L)}\] Molar mass of\[C{{H}_{3}}OH=12+3+16+1=32\]\[g\text{ }mo{{l}^{-1}}\] \[\therefore \] \[0.25\,mol\,{{L}^{-1}}=\frac{Given\text{ }mass}{32\,g\,mo{{l}^{-1}}}\times \frac{1}{4L}\] \[\therefore \]Given mass = 32 g or 0.032 kg Again, \[\because \] \[density=\frac{given\text{ }mass\text{ }(kg)}{volume\text{ }(mL)}\] \[\therefore \] \[0.8\,kg\,{{L}^{-1}}=\frac{0.032\,kg}{V(mL)}\] 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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