Question

A solution was made by putting 350mL of a 2.5%m/v solution of potassium chloride with 250mL of a 0.250mol/L solution of magnesium chloride into a beaker. what is the concentration of chloride ion? (mol/L)

Answer 1

1. A mass of potassium chloride in 2.5% m/v solution found by using formula:

$$2.5 \%m/v=\frac{m KCl}{350mL}\cdot100$$ $$mKCl = \frac {2.5 \%g/mL \cdot 350mL} {100}$$

$$mKCl=8.75 g$$

2. Number of moles of the KCl is found by dividing the mass by molecular weight of this salt (74.55 g/mol):

$$n=\frac{m}{MW}=\frac{8.75g}{74.55g/mol}=0.117moles$$

3. Number of moles of chloride ion in the solution of potassium chloride is:

$$n=1\cdot0.117=0.117moles$$

4. To find a moles of MgCl₂ one must multiply the molar concentration by it's volume in L:

$$n=M \cdot V$$ $$n= 0.250 \frac{mol}{L} \cdot 0.250 L = 0.0625moles$$

5. number of moles of chloride ion in the solution of magneisum chloride is:

$$n=2\cdot 0.0625moles=0.125moles$$

6. Total number of moles of chloride ion in the final solution:

$$n= {0.117moles}+{0.125moles}=0.242 moles$$

7. Total volume of final solution:

$$V= \frac{350mL+250mL}{1000}=0.6L$$

8. Concentration of chloride ion:

$$M=\frac{n}{V}=\frac{0.242moles}{0.6L}=0.4 \frac{moles}{L}$$