JEE MAIN - Chemistry (2021 - 31st August Evening Shift - No. 19)
Sodium oxide reacts with water to produce sodium hydroxide. 20.0 g of sodium oxide is dissolved in 500 mL of water. Neglecting the change in volume, the concentration of the resulting NaOH solution is ______________ $$\times$$ 10$$-$$1 M. (Nearest integer) [Atomic mass : Na = 23.0, O = 16.0, H = 1.0]
Answer
13
Explanation
Na2O + H2O $$\to$$ 2NaOH
$${{20} \over {62}}$$ moles
Moles of NaOH formed = $${{20} \over {62}}$$ $$\times$$ 2
[NaOH] = $${{{{40} \over {62}}} \over {{{500} \over {1000}}}}$$ = 1.29 M = 13 $$\times$$ 10$$-$$1 M (Nearest integer)
$${{20} \over {62}}$$ moles
Moles of NaOH formed = $${{20} \over {62}}$$ $$\times$$ 2
[NaOH] = $${{{{40} \over {62}}} \over {{{500} \over {1000}}}}$$ = 1.29 M = 13 $$\times$$ 10$$-$$1 M (Nearest integer)
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