Question

8. A 50.0 mL 0.05 mol/l solution of sodium cloride (NaCl) was mixed with 100.0 mL

of 0.02 mol/l NaCl solution. What is the mass percent of NaCl in the final solution?
Assume the volumes are additive and their densities 21 g/mL. The molar mass of
NaCl is 58.5 g/mol. (10 points)

Answer 1

`\large \boxed{0.012 \%}`

Step-by-step explanation:

Data:

Solution 1: V₁ = 50.0 mL; c₁ = 0.05 mol·L⁻¹

Solution 2: V₂ = 100 mL; c₂ = 0.02 mol·L⁻¹

NaCl : ρ = 2.1 g/mL

1. Solution 1

(a) Moles of NaCl

`n = \text{50.0 mL} * \frac{\text{0.05 mmol}}{\text{1 L}} = \text{2.5 mmol}`

(b) Mass of NaCl

`m = \text{2.5 mmol} * \frac{\text{58.5 mg}}{\text{1 mmol}} = \text{150 mg} = \text{0.15 g}`

(c) Volume of NaCl

`V = \text{0.15 g} * \frac{\text{1 mL}}{\text{2.1 g}} = \text{0.070 mL}`

(d) Volume of water

V = 50.0 mL - 0.070 mL = 49.9 mL

(e) Mass of water

`\text{Mass} = \text{49.9 mL} * \frac{\text{1.00g}}{\text{1 mL}} = \text{49.9 g}`

2. Solution 2

(a) Moles of NaCl

`n = \text{100 mL} * \frac{\text{0.02 mmol}}{\text{1 mL}} = \text{2.0 mmol}`

(b) Mass of NaCl

`m = \text{2.0 mmol} * \frac{\text{58.5 mg}}{\text{1 mmol}} = \text{120 mg} = \text{0.12 g}`

(c) Volume of NaCl

`V = \text{0.12 g} * \frac{\text{1 mL}}{\text{2.1 g}} = \text{0.053 mL}`

(d) Volume of water

V =100 mL - 0.055 mL = 100 mL

(e) Mass of water

`\text{Mass} = \text{100 mL} * \frac{\text{1.00 g}}{\text{1 mL}} = \text{100 g}`

3. Combined solutions

(a) Mass of NaCl

Mass of NaCl = 0.015 g + 0.012 g = 0.018 g

(b) Mass of water

Mass of water = 49.9 g + 100 g = 150 g

(c) Mass percent

`\text{Mass percent} = \frac{\text{Mass of component}}{\text{Total mass}} * \, 100 \,\%\\\\\text{Mass \% NaCl} = \frac{\text{0.018 g}}{\text{150 g}}* \, 100 \% = \mathbf{0.012 \%}\\\\\text{The mass percent of NaCl is $\large \boxed{\mathbf{0.012 \, \%}}$}`