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

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