Some Basic Concepts of Chemistry - Result Question 6
6. 50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
(2019 Main, 12 Jan I)
(a) 40 g
(b) 80 g
(c) 20 g
(d) 10 g
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Answer:
Correct Answer: 6. (*)
Solution: The reaction takes place as follows,
$ \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{H}_2 \mathrm{O} $
Now, 50 mL of $0.5 \mathrm{M} \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$ is needed to neutralize 25 mL of NaOH .
$\therefore$ Meq of $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4=$ Meq of NaOH
$ \begin{aligned} 50 \times 0.5 \times 2 & =25 \times M_{\mathrm{NaOH}} \times 1 \\ M_{\mathrm{NaOH}} & =2 \mathrm{M} \end{aligned} $
Now,
$ \begin{aligned} \text { molarity } & =\dfrac{\text { Number of moles }}{\text { Volume of solution (in L) }} \\ & =\dfrac{\text { Weight } / \text { molecular mass }}{\text { Volume of solution (in L) }} \\ 2 & =\dfrac{w_{\mathrm{NaOH}}}{40} \times \dfrac{1000}{50} \\ w_{\mathrm{NaOH}} & =\dfrac{2 \times 40 \times 50}{1000}=4 \mathrm{~g} \end{aligned} $
Thus, $\left({ }^*\right)$ none option is correct.
BETA
