Question

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.

Answer #1

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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