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 (integer)
15
Solution
<p>$$\mathrm{\mathop {Na + {H_2}O}\limits_{0.69\,g} \to \mathop {NaOH + {1 \over 2}{H_2}}\limits_{0.03\,moles}} $$</p>
<p>= 0.03 moles</p>
<p>$\therefore 0.03=2\times\mathrm{V}$</p>
<p>$\mathrm{V=\frac{0.03}{2}L}$</p>
<p>= 15 mL</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
This question is part of PrepWiser's free JEE Main question bank. 163 more solved questions on Some Basic Concepts of Chemistry are available — start with the harder ones if your accuracy is >70%.