Given 0.02% v/v (percent by volume) solution of methanol in water, convert the number to % mass, mg/L and Molarity. Density of methanol is 791 g/L and density of water is 997 g/L at 25 °C.
Ans :
Let the volume of solution be 100 mL
So 0.02% v/v solution will have :
0.02 = (v / 100) x 100
= 0.02 mL of methanol
Volume of water will be = 100 - 0.02 = 99.98 mL
Density = mass / volume
791g/L or 0.791 g/mL = m / 0.02
mass of methanol = 0.01582 g
mass of water = 99.98 x 0.997 = 99.68 g
Mass of solution = 0.01582 + 99.68 = 99.696 g
% mass = ( mass of solute / mass of solution) x 100
= (0.01582 / 99.696) x 100
= 0.016 %
Mass of methanol = 0.01582 g or 15.82 mg
Volume = 100 mL or 0.1 L
So concentration in mg/L = 15.82 / 0.1
= 158.2 mg/L
Number of moles of methanol = 0.01582 / molar mass
= 0.01582 / 32.04 = 4.94 x 10-4 mol
Molarity = no. of moles of solute / volume of solution in L
= (4.94 x 10-4) / 0.1
= 4.94 x 10-3 M
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