Question

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

Answer 1

Answer:

a. 74.55 g/mol.

b. 91.70 g.

c. 8.40%.

d. 1.23 mol/L.

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.