1 g methanol ( d =0.75 g mL -1) is mixed with 1 g water ( d =1.00 g mL -1) at 298 K. Thus, volume % of methanol ( V / V ) is

Question

1 g methanol ( d =0.75 g mL -1) is mixed with 1 g water ( d =1.00 g mL -1) at 298 K. Thus, volume % of methanol ( V / V ) is (A) 42.86 % (B) 57.14 % (C) 50.00 % (D) 25.00 %

Tardigrade Admin · Accepted Answer

Methanol =1 g =(1/0.75) mL =(100/75) mL H 2 O =1 g =(1/1) mL =(75/75) mL Total volume of mixture =(100/75)+(75/75)=(175/75) ∴ % methanol =(((100/75)/(175)75)) × 100=(400/7)=57.14 %

Q. $1 g$ methanol $(d = 0.75 g m L^{- 1})$ is mixed with $1 g$ water $(d = 1.00 g m L^{- 1})$ at $298 K$ . Thus, volume $%$ of methanol $(V / V)$ is

$42.86%$

$57.14%$

$50.00%$

$25.00%$

Methanol $= 1 g = \frac{1}{0.75} m L = \frac{100}{75} m L$

$H_{2} O = 1 g = \frac{1}{1} m L = \frac{75}{75} m L$

Total volume of mixture $= \frac{100}{75} + \frac{75}{75} = \frac{175}{75}$

$∴ %$ methanol $= (\frac{\frac{100}{75}}{\frac{175}{75}}) \times 100 = \frac{400}{7} = 57.14%$

Q. 1g methanol (d=0.75gmL−1) is mixed with 1g water (d=1.00gmL−1) at 298K. Thus, volume % of methanol (V/V) is

42.86%

57.14%

50.00%

25.00%