Question

A solution is prepared by dissolving 23.7 g of CaCl 2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of Cl- in this solution is ________ M.

Also, The concentration of HCl in a solution that is prepared by dissolving 5.5 g of HCl in 200 g of C2H6O is ________ molal.

Answer 1

Ans 1 :Number of moles of CaCl2 = 23.7 / molar mass

= 23.7 / 110.98

= 0.2135 mol

So number of moles of Cl- = 2 x 0.2135

= 0.4271 mol

Mass of solution = 23.7 + 375 = 398.7 g

Density = mass / volume

volume = 398.7 / 1.05

= 379.7 mL

Molarity = no. of moles of Cl- / volume of solution in L

= 0.4271 / 0.3797

= 1.12 M

Ans 2 :

Numbe rof moles of HCl = 5.5 / molar mass

= 5.5 / 36.46

= 0.15 mol

Molality = no. of moles of HCl / mass of solvent in kg

= 0.15 / 0.200

= 0.75 molal