Question

The molarity of solution containing 15 g of NaOH in 900 ml solution will be ______.

• A. 0.42 M
• B. 0.042 M
• C. 4.2 M
• D. 42.0 M

Hint:

For molarity questions, always use the formula \(\displaystyle M=\frac{\text{moles of solute}}{\text{volume in litres}}\). Also remember to convert mL into L before substituting in the formula.

Answer 1

The Correct Option is A

Step 1: Write the formula of molarity.
Molarity is defined as the number of moles of solute present in \(1\) litre of solution. Therefore,
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}} \] Here, the solute is \(\mathrm{NaOH}\), its mass is \(15\) g, and the volume of solution is \(900\) mL.
Step 2: Calculate the molar mass of \(\mathrm{NaOH}\).
The molar mass of sodium hydroxide is calculated as follows:
\[ \mathrm{NaOH} = \mathrm{Na} + \mathrm{O} + \mathrm{H} = 23 + 16 + 1 = 40 \, \mathrm{g/mol} \] So,
\[ \text{Molar mass of NaOH} = 40 \, \mathrm{g/mol} \] Step 3: Calculate the number of moles of \(\mathrm{NaOH}\).
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{15}{40} = 0.375 \, \text{mol} \] Thus, the number of moles of \(\mathrm{NaOH}\) present in the solution is \(0.375\) mol.
Step 4: Convert volume into litres and find molarity.
Given volume of solution \(= 900\) mL
Since,
\[ 1000 \, \text{mL} = 1 \, \text{L} \] therefore,
\[ 900 \, \text{mL} = \frac{900}{1000} = 0.9 \, \text{L} \] Now,
\[ \text{Molarity} = \frac{0.375}{0.9} = 0.4167 \, \text{M} \] Approximating to two decimal places,
\[ \text{Molarity} \approx 0.42 \, \text{M} \] Step 5: Conclusion.
Hence, the molarity of the given \(\mathrm{NaOH}\) solution is \(0.42 \, \text{M}\). Therefore, the correct option is \((A)\).
Final Answer:0.42 M.