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

`\boxed{\text{20.2 mL}}`

Step-by-step explanation:

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

1. Mass of stock solution

`\text{Mass} = \text{1000 mL} * \frac{\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} * \frac{\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)* \frac{\text{1 mol NH}_(3)}{\text{17.03 g NH}_(3)}= \text{14.89 mol NH}_(3)`

4. Molar concentration of stock solution

`c = \frac{\text{moles}}{\text{litres}} = \frac{\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 * 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.}`