Question

In a laboratory experiment, a 14.0 mL sample of KCl solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and KCl solution is 44.30 g. After heating, the evaporating dish and dry KCl have a combined mass of 27.90 g.
(a) What is the mass percent (m/m) of the KCl solution?
(b) What is the molarity ( M) of the KCl solution?
(c) If water is added to 10.0 mL of the initial KCl solution to give a final volume of 60.0 mL, what is the molarity of the diluted KCl solution?

Answer 1

Answer:

a) 18.81% is the mass percent (m/m) of the KCl solution.

b) 3.643 M is the molarity of the KCl solution.

c) 0.6073 M is the molarity of the diluted KCl solution.

Explanation:

a) mass on of evaporating dish = 24.10 g = x

mass on of evaporating dish and KCl solution = 44.30 g = y

After heating, mass on of evaporating dish and dry KCl  = 27.90 g = z

Mass of KCl solution = y - x = 44.30 g - 24.10 g = 20.2 g

Mass of KCl in solution = z - x = 27.90 g - 24.10 g = 3.8 g

The mass percent (m/m) of the KCl solution;

$=\frac{\text{Mass of KCl}}{\text{Mass of KCL solution}}\times 100$

$=\frac{3.8 g}{20.2 g}\times 100=18.81\%$

18.81% is the mass percent (m/m) of the KCl solution.

b) Moles of KCl = $\frac{3.8 g}{74.5 g/mol}=0.05101 mol$

Volume of the KCl solution = 14.0 mL = 0.014 L ( 1 mL = 0.001 L)

Molarity of the solution :

$\frac{0.05101 mol}{0.014 L}=3.644 M$

3.643 M is the molarity of the KCl solution.

c)

Molarity of KCl solution before dilution = $M_1=3.644 M$

Volume of the solution before dilution = $V_1=10 mL$

Molarity of KCl solution after dilution = $M_2=?$

Volume of the solution after dilution = $V_2=60.0 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{3.644M\times 10.0 mL}{60.0mL}=0.6073 M$

0.6073 M is the molarity of the diluted KCl solution.