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प्रश्न
Sulphuric acid reacts with sodium hydroxide as follows:
\[\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}\]
When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is:
(i) 0.1 mol L–1
(ii) 7.10 g
(iii) 0.025 mol L–1
(iv) 3.55 g
उत्तर
(ii) 7.10 g
(iii) 0.025 mol L–1
Explanation:
0.1 mole \[\ce{H2SO4}\] reacts with 1 mole of \[\ce{NaOH}\].
0.1 mole of \[\ce{NaOH}\] will react with `0.1/2` mole of \[\ce{H2SO4}\].
Here, \[\ce{NaOH}\] is the limiting reagent.
2 mole of \[\ce{NaOH}\] produces 1 mole of \[\ce{Na2SO4}\].
0.1 mole of \[\ce{NaOH}\] will give `0.1/2` mole of \[\ce{Na2SO4}\].
No. of mole = `"Given mass"/("Molar mass" (Na_2SO_4))`
On substituting the value in tha above equation, the mass can be calculated as 0.05 mol = `"Given mass"/(142 g mol^-1)`
Given mass = 7.10 g
Volume of solution after mixing is 2 L.
So, the molarity of \[\ce{Na2SO4}\] is,
Molarity = `0.05/2` = 0.025 mol L–1
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