Question

$5 \,g$ of $NaOH$ was dissolved in deionized water to prepare a $450 \,mL$ stock solution What volume (in $mL$ ) of this solution would be required to prepare $500 \,mL$ of $01\, M$ solution? Given: Molar Mass of $Na, O$ and $H$ is 23,16 and $1 \,g\, mol ^{-1}$ respectively

Answer 1

Correct Answer: 180

The correct answer is 180.
To prepare 500 mL of 0.1 M NaOH solution, we need to calculate the volume of the stock solution required.
First, let's calculate the number of moles of NaOH in 500 mL of 0.1 M solution:
Number of moles of NaOH = Molarity × Volume (in L) Number of moles of NaOH = 0.1 × 0.5 Number of moles of NaOH = 0.05

Next, let's calculate the number of moles of NaOH in the stock solution:
Number of moles of NaOH in stock solution = (mass of NaOH / molar mass of NaOH) = (5 g / 40 g/mol) = 0.125 mol
Now, we can use the following formula to calculate the volume of the stock solution required:
Volume of stock solution = (Number of moles required / Number of moles in stock solution) × Volume of stock solution
Plugging in the values, we get:
Volume of stock solution = (0.05 / 0.125) × 0.45 L Volume of stock solution = 0.18 L = 180 mL

Therefore, 180 mL of the NaOH stock solution is required to prepare 500 mL of 0.1 M NaOH solution.

Answer 2

The correct answer is 180
\(M=\frac{5}{40} \times \frac{1000}{450}\)

\(M_1V_1=M_2V_2\)

\((\frac{5}{40} \times \frac{1000}{450})\times V_1=0.1 \times 500\)

\(V_1=180\)
\(\therefore\) volume required is 180