Question

A stock solution is 28.2 percent ammonia (NH3) by mass, and the solution has a density of 0.8990 grams per milliliter. What volume, in milliliters, of this stock solution is required to prepare 600 milliliters of a 0.500 molar ammonia solution?

Answer 1

Answer:

$\boxed{\text{20.2 mL}}$

Explanation:

Assume that the volume of the stock solution is 1 L.

1. Mass of  stock solution

$\text{Mass} = \text{1000 mL} \times \dfrac{\text{0.8990 g}}{\text{1 mL}} = \text{899.0 g}$

2.Mass of NH₃

$\text{Mass of NH}_{3} = \text{899.0 g stock} \times \dfrac{\text{28.2 g NH}_{3}}{\text{100 g stock}} = \text{253.5 g NH}_{3}$

3. Moles of NH₃

$\text{Moles of NH}_{3} = \text{253.5 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.89 mol NH}_{3}$

4. Molar concentration of stock solution

$c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{14.89 mol}}{\text{1 L}} = \text{14.89 mol/L}$

5. Volume of stock needed for dilution

Now that you know the concentration of the stock solution, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

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:

$\begin{array}{rcl}14.89V_{1} & = & 0.500 \times 600\\14.89V_{1} & = & 300\\V_{1} & = & \text{20.2 mL}\\\end{array}\\\text{You will need \boxed{\textbf{20.2 mL}} of the stock solution.}$