Sankalp NEET Full Test-1 Question-79 Solution

Question: 79. 1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCL solution, the mass of sodium hydroxide left unreacted is equal to

(1) 750 mg

(2) 250 mg

(3) Zero mg

(4) 200 mg

Answer: Option (2)

Explanation:

The balanced chemical reaction between sodium hydroxide and hydrochloric acid is

\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O}

Number of moles of \mathrm{NaOH} initially present is

\text{Moles of NaOH}=\dfrac{1}{40} = 0.025

Number of moles of \mathrm{HCl} used is

\text{Moles of HCl} = 0.75 \times \dfrac{25}{1000} = 0.01875

From the reaction equation, \mathrm{NaOH} and \mathrm{HCl} react in a 1:1 molar ratio.

Thus, 0.01875 moles of \mathrm{NaOH} will react completely with \mathrm{HCl}.

Moles of \mathrm{NaOH} left unreacted are

0.025 - 0.01875 = 0.00625

Mass of unreacted \mathrm{NaOH} is

0.00625 \times 40 = 0.25 \text{ g} 0.25 \text{ g} = 250 \text{ mg}

Therefore, the mass of sodium hydroxide left unreacted is 250 \text{ mg}.

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