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

Correct Answer: 7

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