Sulphuric acid reacts with sodium hydroxide as follows :
H2SO4 + 2NaOH → Na2SO4 + 2H2O
When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained
is
(i) 1 mol L–1
(ii) 10 g
(iii) 025 mol L–1
(iv) 55 g
Sulphuric acid reacts with sodium hydroxide as follows :
H2SO4 + 2NaOH → Na2SO4 + 2H2O
When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is
(i) 1 mol L–1
(ii) 10 g
(iii) 025 mol L–1
(iv) 55 g
-
1 Answer
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This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (B), (C)
0.1 mole H2SO4 reacts with 1 mole of NaOH.
0.1 mole of NaOH will react with = mole of H2SO4
Here, NaOH is the limiting reagent.
2 mole of NaOH produces 1 mole of Na2SO4
0.1 mole of NaOH will give mole of Na2SO4
No. of mole =
On substituting the value in the above equation, the mass can be calculated as
0.05 mol =
given mass = 7.10 g
Volume of solution after mixing is 2 L.
So, the molarity of Na2SO4 is
Molarity = = 0.025mol L-1
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