Answer:
[tex]\large \boxed{0.012 \%}[/tex]
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
[tex]n = \text{50.0 mL} \times \dfrac{\text{0.05 mmol}}{\text{1 L}} = \text{2.5 mmol}[/tex]
(b) Mass of NaCl
[tex]m = \text{2.5 mmol} \times \dfrac{\text{58.5 mg}}{\text{1 mmol}} = \text{150 mg} = \text{0.15 g}[/tex]
(c) Volume of NaCl
[tex]V = \text{0.15 g} \times \dfrac{\text{1 mL}}{\text{2.1 g}} = \text{0.070 mL}[/tex]
(d) Volume of water
V = 50.0 mL - 0.070 mL = 49.9 mL
(e) Mass of water
[tex]\text{Mass} = \text{49.9 mL} \times \dfrac{\text{1.00g}}{\text{1 mL}} = \text{49.9 g}[/tex]
2. Solution 2
(a) Moles of NaCl
[tex]n = \text{100 mL} \times \dfrac{\text{0.02 mmol}}{\text{1 mL}} = \text{2.0 mmol}[/tex]
(b) Mass of NaCl
[tex]m = \text{2.0 mmol} \times \dfrac{\text{58.5 mg}}{\text{1 mmol}} = \text{120 mg} = \text{0.12 g}[/tex]
(c) Volume of NaCl
[tex]V = \text{0.12 g} \times \dfrac{\text{1 mL}}{\text{2.1 g}} = \text{0.053 mL}[/tex]
(d) Volume of water
V =100 mL - 0.055 mL = 100 mL
(e) Mass of water
[tex]\text{Mass} = \text{100 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100 g}[/tex]
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
[tex]\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100 \,\%\\\\\text{Mass \% NaCl} = \dfrac{\text{0.018 g}}{\text{150 g}}\times \, 100 \% = \mathbf{0.012 \%}\\\\\text{The mass percent of NaCl is $\large \boxed{\mathbf{0.012 \, \%}}$}[/tex]
