Question

An aqueous solution containing 50% K2CO3 has a density of 1.5404 g/mL.

a) how many grams of K2CO3 are contained in 1 L of solution.

b) Calculate the molarity (mol/L) of the K2CO3 solution

c) Calculate the molarity (mol/kg) of the K2CO3 solution.

Answer 1

a)

consider 1 L of solution

so,

volume of solution = 1 L = 1000 mL

density = 1.5404 g/mL

So,

mass of solution = density of solution * volume of solution

= 1.5404 g/mL * 1000 mL

= 1540.4 g

mass of K2CO3 = 50% of mass of solution

= 50*1540.4/100

= 770.2 g

Answer: 770.2 g

b)

Molar mass of K2CO3,

MM = 2*MM(K) + 1*MM(C) + 3*MM(O)

= 2*39.1 + 1*12.01 + 3*16.0

= 138.21 g/mol

mass(K2CO3)= 770.2 g

number of mol of K2CO3,

n = mass of K2CO3/molar mass of K2CO3

=(770.2 g)/(138.21 g/mol)

= 5.5727 mol

Molarity = mol of solute / volume of solution in L

= 5.5727 mol / 1 L

= 5.5727 M

Answer: 5.5727 M

c)

mass of solvent = mass of solution - mass of K2CO3

= 1540.4 g - 770.2 g

= 770.2 g

= 0.7702 Kg

Molality = mol of solute / mass of solvent in Kg

= 5.5727 mol / 0.7702 Kg

= 7.235 mol/Kg

Answer: 7.235 mol/Kg