Q.

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

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a

42.86%

b

57.14%

c

50.00%

d

25.00%

answer is B.

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Detailed Solution

Methanol =1g=10.75mL=10075mLH2O=1g=11mL=7575mLTotal volume of mixture=10075+7575=17575∴% methanol =1007517575×100=4007=57.14%
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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