Answer:
[tex]\boxed{\text{20.2 mL}}[/tex]
Explanation:
Assume that the volume of the stock solution is 1 L.
1. Mass of stock solution
[tex]\text{Mass} = \text{1000 mL} \times \dfrac{\text{0.8990 g}}{\text{1 mL}} = \text{899.0 g}[/tex]
2.Mass of NH₃
[tex]\text{Mass of NH}_{3} = \text{899.0 g stock} \times \dfrac{\text{28.2 g NH}_{3}}{\text{100 g stock}} = \text{253.5 g NH}_{3}[/tex]
3. Moles of NH₃
[tex]\text{Moles of NH}_{3} = \text{253.5 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.89 mol NH}_{3}[/tex]
4. Molar concentration of stock solution
[tex]c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{14.89 mol}}{\text{1 L}} = \text{14.89 mol/L}[/tex]
5. Volume of stock needed for dilution
Now that you know the concentration of the stock solution, you can use the dilution formula .
[tex]c_{1}V_{1} = c_{2}V_{2}[/tex]
to calculate the volume of stock solution.
Data:
c₁ = 14.89 mol·L⁻¹; V₁ = ?
c₂ = 0.500 mol·L⁻¹; V₂ = 600 mL
Calculations:
[tex]\begin{array}{rcl}14.89V_{1} & = & 0.500 \times 600\\14.89V_{1} & = & 300\\V_{1} & = & \text{20.2 mL}\\\end{array}\\\text{You will need $\boxed{\textbf{20.2 mL}}$ of the stock solution.}[/tex]
