MHT CET · Chemistry · Solutions
If \(2 \cdot 0 \mathrm{~g}\) of \(\mathrm{NaOH}\) is dissolved in \(500 \mathrm{~cm}^{3}\) of water, what is molarity of solution?
- A \(0 \cdot 25 \mathrm{~mol} \mathrm{dm}^{-3}\)
- B \(0 \cdot 1 \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(0 \cdot 4 \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(0 \cdot 50 \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(B) \(0 \cdot 1 \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
Molarity \(=0.10 \cdot \mathrm{mol} \cdot \mathrm{L}^{-1}\) with respect to \(\mathrm{NaOH}\)
Explanation: Molarity \(=\frac{\text { Moles of solute }}{\text { Volume of solution }}\)
Molarity \(=\frac{2 \cdot g}{40.0 \cdot g \cdot m o l^{-1}} \times \frac{1}{500 \cdot m L \times 10^{-3} \cdot L \cdot m L^{-1}}\)
Molarity \(=0.1 \mathrm{~mol} \mathrm{dm}^{\wedge}-3\)
Explanation: Molarity \(=\frac{\text { Moles of solute }}{\text { Volume of solution }}\)
Molarity \(=\frac{2 \cdot g}{40.0 \cdot g \cdot m o l^{-1}} \times \frac{1}{500 \cdot m L \times 10^{-3} \cdot L \cdot m L^{-1}}\)
Molarity \(=0.1 \mathrm{~mol} \mathrm{dm}^{\wedge}-3\)
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