Question

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you need to prepare the 0.1M solution?

Answer 1

`\boxed{\text{3.3 mL}}`

Step-by-step explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

`\text{Mass of NaOH} = \text{1000 mL solution } * \frac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}`

2. Moles of NaOH

`\text{Moles of NaOH} = \text{300 g NaOH} * \frac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}`

3. Molar concentration of NaOH

`c= \frac{\text{moles}}{\text{litres}} = \frac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}`

4. Volume of NaOH

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

`c_(1)V_(1) = c_(2)V_(2)`

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

`V = \text{250 mL} * \frac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}`

(b) Calculate the volume of dilute solution

`\begin{array}{rcl}7.50V_(1) & = & 0.1 * 0.250\\7.50V_(1) &= & 0.0250\\V_(1) & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}`

`\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}`