A sample of rubbing alcohol solution contains ethanol, C2H5OH(l). If the sample contains 1.25x10^23 atoms of hydrogen in the ethanol, what amount in moles of ethanol is in the sample?

Mass of C2H5OH = 1.25 g Mass of H2O = 11.6 g Mass percent of C2H5 OH = 1.25 × 100 / 1.25 + 11.6 = 125 / 12.86...

Volume/volume percent = x 100 %Volume of solute =volume of solute Volume of solution Volume/volume % x volume of solution 100%{70% x 500 m L}÷100%= 350 m L350 m L...

a) Molality of the alcohol =8.6Mb)Weight percent of alcohol in the solution=28.35%

Depression in freezing point is denoted by,Freezing point of water + Freezing point of solutionFrom the given dataButWhere freezing point depression constant for water

21,91 mole

3.25 x 105 = 325 000

n(CH3CHOHCH3) = m(CH3CHOHCH3)/m(CH3CHOHCH3) = 10.0g/60g/mol = 0.1667mol.Q = L(CH3CHOHCH3) * n(CH3CHOHCH3) = 42.1 kJ/mol * 0.1667mol = 7.02kJ.

0.28 x 0.79 = 0.22 g.0.22/46 = 0.004 .

According to the equation we obtain 3 molecules of H2O from each molecule of C2H5OH. Thus, from 0.100 mole of C2H5OH we will obtain 0.300 mole of H2O.

197 g