Question

2. A solution is made by adding 1.23 mol of KCl to 1000.0 g of water. Assume that the

density of water (and the resulting solution) is 1.00 g/mL.
a. Calculate the formula weight of KCl.
b. Calculate the mass of KCl in grams.
c. Calculate the percent by mass of KCl in this solution.
d. Calculate the molarity of the solution.

Answer 1

a. 74.55 g/mol.

b. 91.70 g.

c. 8.40%.

d. 1.23 mol/L.

Step-by-step explanation:

a. Calculate the formula weight of KCl.

∵ Formula weight of KCl = atomic weight of K + atomic weight of Cl

atomic weight of K = 39.098 g/mol, atomic weight of Cl = 35.45 g/mol.

∴ Formula weight of KCl = atomic weight of K + atomic weight of Cl = 39.098 g/mol + 35.45 g/mol = 74.548 g/mol ≅ 74.55 g/mol.

b. Calculate the mass of KCl in grams.

• we can use the relation:

no. of moles (n) = mass/molar mass.

∴ mass of KCl = n*molar mass = (1.23 mol)*(74.55 g/mol) = 91.69 g ≅ 91.70 g.

c. Calculate the percent by mass of KCl in this solution.

The mass % of KCl = (mass of KCl/mass of the solution) * 100.

mass of KCl = 91.70 g,

mass of the solution = 1000.0 g of water + 91.70 g of KCl = 1091.70 g.

∴ The mass % of KCl = (91.70 g/1091.70 g)*100 = 8.399% ≅ 8.40%.

d. Calculate the molarity of the solution.

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of KCl)/(Volume of the solution (L))

no. of moles of KCl = 1.23 mol,

Volume of the solution = mass of water / density of water = (1000.0 g)/(1.00 g/mL) = 1000.0 mL = 1.0 L.

M = (1.23 mol)/(1.0 L) = 1.23 mol/L.