Question

Number of moles of NaOH required for the complete neutralisation of 50 mL 0.1 M HCl is
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Answer 1

Answer:

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Answer 2

Answer:

The number of moles of $NaOH$ needed for the complete neutralisation of  $0.1 M HCl$ measured is $5\times 10^{-3}mol$.

Therefore, option a) $5\times 10^{-3}mol$ is the correct choice.

Explanation:

Given,

The volume of $HCl$, V = $50 ml$ = $0.05L$

The molarity of $HCl$, M = $0.1M$

The number of moles of $NaOH$ required for complete neutralization of $HCl$ =?

Now,

• The given neutralization reaction has shown below:
• $NaOH + HCl \rightarrow NaCl + H_{2}O$

From the reaction, we can see that the same number of moles of $NaOH$ is required to neutralize $HCl$.

Therefore,

• Number of moles of $NaOH$ = number of moles of $HCl$

Now, as we know,

• Molarity = $\frac{Number of moles}{Volume}$
• Number of moles = molarity × volume = $0.05 \times 0.1$ = $5\times 10^{-3} mol$.

Hence, the number of moles of $NaOH$ required to neutralize $HCl$ = $5\times 10^{-3} mol$.