Question

6.3 g of MgCl2 is dissolved in 200 mL of water. What is the resulting concentration of Cl–(aq) ions?



The atomic weigth of Mg is 24.3, that of Cl is 35.5. Give your answer in moles·L–1 and in 2 decimals.

How do I work this out?

Answer 1

Answer on Question #43439 - Chemistry - Inorganic Chemistry

Question:

6.3 g of MgCl₂ is dissolved in 200 mL of water. What is the resulting concentration of Cl⁻ (aq) ions?

The atomic weight of Mg is 24.3, that of Cl is 35.5. Give your answer in mol·L⁻¹ and in 2 decimals.

Solution:

MgCl₂ dissociates as follows:

$$\mathrm{MgCl_2} \leftrightarrow \mathrm{Mg^{2+}} + 2\mathrm{Cl^-}$$

We see that each molecule of magnesium chloride dissociates producing 2 Cl⁻ ions, i.e. chloride ions' concentration is twice as big as that of magnesium chloride.

Molar concentration of MgCl₂ can be found from the expression:

$$C_M = \frac{n}{V} = \frac{m}{MV}$$

- n – Number of moles of MgCl₂;

- V – The volume of the solution, V = 200 mL = 0.2 L;

- m – The mass of MgCl₂ dissolved, m = 6.3 g;

- M – Molar mass of MgCl₂, M = M(Mg) + 2M(Cl) = 24.3 + 2·35.5 = 95.3 g/mol.

$$C_M(MgCl_2) = \frac{6.3}{95.3 \cdot 0.2} = 0.33\ \mathrm{mol/L}$$

So, the resulting concentration of Cl⁻(aq) ions is:

$$C_M(Cl^-) = 2C_M(MgCl_2) = 2 \cdot 0.33 = 0.66\ \mathrm{mol/L}$$

Answer: 0.66 mol/L.

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