Question

\text{Volume of } 3 \, \text{M NaOH (formula weight 40 g mol}^{-1}\text{) which can be prepared from } 84 \, \text{g of NaOH is } \\_\\_\\_\\_\\_ \times 10^{-1} \, \text{dm}^3.

Answer 1

The molarity formula is given by: $M = \frac{n_{\text{NaOH}}}{V_{\text{sol}}},$

where:
- $M$ is the molarity (in mol/L),
- $n_{\text{NaOH}}$ is the number of moles of NaOH,
- $V_{\text{sol}}$ is the volume of the solution (in liters).

Given:
- $M = 3 \, \text{M}$,
- Mass of NaOH = $84 \, \text{g}$,
- Molar mass of NaOH = $40 \, \text{g/mol}$.

Step 1: Calculate the number of moles of NaOH

$n_{\text{NaOH}} = \frac{\text{Mass of NaOH}}{\text{Molar mass}} = \frac{84}{40} = 2.1 \, \text{moles}.$

Step 2: Calculate the volume of the solution using the molarity formula

$V_{\text{sol}} = \frac{n_{\text{NaOH}}}{M} = \frac{2.1}{3} = 0.7 \, \text{L}.$

Expressing the volume in scientific notation:

$V_{\text{sol}} = 7 \times 10^{-1} \, \text{L}.$

Final Answer: $V_{\text{sol}} = 0.7 \, \text{L or } 7 \times 10^{-1} \, \text{L}.$

The Correct answer is: 7