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Allushta [10]
3 years ago
9

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 n

eed to prepare the 0.1M solution?
Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:

\boxed{\text{3.3 mL}}

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 } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}

2. Moles of NaOH  

\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}

3. Molar concentration of NaOH

c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\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} \times \dfrac{ \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 \times 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.}

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