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 [2024]

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Solution

Q.
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 [2024]

1 750 mg

2 250 mg

3 zero mg

4 200 mg

Ans.
(2)

Molarity $= \frac{Moles}{Volume (mL)} \times 1000$

For HCl,

Number of moles $= \frac{M \times V (mL)}{1000} = \frac{0.75 \times 25}{1000} = \frac{0.75}{40}$

Mass of HCl reacted $= \frac{0.75}{40} \times 36.5 = 0.684$ $g$

According to balanced equation

$\underset{1 mole (40 g)}{NaOH}$ $+$ $\underset{1 mole (36.5 g)}{HCl}$ $\to NaCl + H_{2} O$

36.5 g HCl reacts with 40 g NaOH

0.684 g HCl will react with $\frac{40}{36.5} \times 0.684 = 0.75$ $g$

NaOH left unreacted $= 1 - 0.75 = 0.25 g or 250 mg$

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