Question

148.2 g of Cupric sulfate are dissolved in enough water to make 2.00 x 103 mL of total solution. What is the molar concentration?

When the same amount of cupric sulfate from problem 1 is dissolved in 1,375 g of water, what is the molal concentration of the resulting solution?

Answer 1

Task:

148.2 g of Cupric sulfate are dissolved in enough water to make 2.00 x 10³ mL of total solution. What is the molar concentration?

When the same amount of cupric sulfate from problem 1 is dissolved in 1,375 g of water, what is the molal concentration of the resulting solution?

Solution:

Problem 1.

The molarity of solution is

$$C (M) = n (mol) / V(L)$$

C – molarity (M)

n – number of moles of substance

V – volume of solution (L)

First of all we have to convert mL to L

$$1 \text{ L} = 1.0 \cdot 10^{3} \text{ mL}$$

The volume of solution is

$$V = 2.00 \cdot 10^{3} \text{ mL} = 2.00 \text{ L}$$

The mass of CuSO₄ is

$$m(g) = n(mol) \cdot MW(g/mol)$$

The molar weight consists of atomic weights of elements taken from the periodic table

$$MW (CuSO_4) = MW (Cu) + MW(S) + 4 \cdot MW (O) = 63.5 + 32 + 3 \cdot 16 = 159.5 \text{ g/mol}$$

The number of moles of CuSO₄

$$n(mol) = m(g) / MW(g/mol)$$ $$n(mol) = 148.2 / 159.5 = 0.929 \text{ mol}$$

The molarity is

$$C (M) = n (mol) / V(L)$$ $$C (CuSO_4) = 0.929 / 2.00 = 0.46 \text{ M}$$

Problem 2.

The molal concentration is the amount of substance (mol) in 1 kg of solvent.

$$M = n(mol) / m(kg)$$

We have to convert g to kg

1 kg = 1.0 · 10³ g

The mass of solvent (water) is

m (kg) = 1,375 / 1.0 · 10³ = 1.375 kg

The molality of solution is

M = 0.929 / 1.375 = 0.676 mol/kg

Answer: 1) C (CuSO₄) = 0.46 M

2) M (CuSO₄) = 0.676 mol/kg