Question

Sol. [Key Concept: Limiting reagent in neutralisation]

$$M = \frac{W \times 1000}{M_{2} \times V(inmL)}$$

$$\begin{matrix} & W = \frac{M \times M_{2} \times V(\text{~in~}mL)}{1000} = \frac{0.75 \times 36.5 \times 25}{1000} \\ & \ = 0.684\text{ }g(\text{~Mass of~}HCl) \\ & \underset{36.5\text{ }g}{HCl} + \underset{40\text{ }g}{NaOH} \longrightarrow HCl + NaOH \end{matrix}$$

36.5 g HCl reacts with $NaOH = 40\text{ }g$

0.684 g HCl reacts with $NaOH = \frac{40}{36.5} \times 0.684 \simeq 0.750\text{ }g$

Amount of NaOH left $= 1\text{ }g - 0.750\text{ }g = 0.250\text{ }g = 250mg$