JEE MAIN - Chemistry (2023 - 29th January Evening Shift - No. 16)
The volume of HCl, containing 73 g L$$^{-1}$$, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is __________ mL. (Nearest Integer)
(Given : molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g mol$$^{-1}$$ respectively.)
Answer
15
Explanation
$$\mathrm{\mathop {Na + {H_2}O}\limits_{0.69\,g} \to \mathop {NaOH + {1 \over 2}{H_2}}\limits_{0.03\,moles}} $$
= 0.03 moles
$$\therefore 0.03=2\times\mathrm{V}$$
$$\mathrm{V=\frac{0.03}{2}L}$$
= 15 mL
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