Question

You wish to prepare 199 grams of 11.8 % MgCl₂.

You will need ___grams of magnesium chloride and ______ mL of water.

Assume that the density of water is 1.00 g / ml.

Answer 1

We have, Percentage by mass of solute =[ (Mass of solute ) / ( Mass of solute + Mass of solvent ) ] x 100

Percentage by mass of MgCl₂ = [ Mass of MgCl₂ / 199 g ] x 100 = 11.8

Mass of MgCl₂ = 11.8 /100 x 199 = 23.48 g

Mass of solution = Mass of Solute + Mass of water

Mass of water = Mass of solution - Mass of MgCl₂ = 199 g - 23.48 g = 175.5 g

We have density of water = Mass of water / Volume of water = 1 g/ ml.

Therefore, Volume of water = Mass of water / density = 175.5 g / ( 1 g/ml) = 175.5 ml

ANSWER: Mass of MgCl₂ = 23.48 g and volume of water = 175.5 ml