JEE MAIN - Chemistry (2020 - 9th January Morning Slot - No. 13)
The molarity of HNO3 in a sample which has
density 1.4 g/mL and mass percentage of 63%
is _____.
(Molecular Weight of HNO3 = 63)
(Molecular Weight of HNO3 = 63)
Answer
14
Explanation
%w/w = 63%
dsolution = 1.4 g/ml
Molarity =
= $${{63 \times 1.4 \times 10} \over {63}}$$ = 14 M
dsolution = 1.4 g/ml
Molarity =
%w/w $$ \times $$ dsolution
Moleculer weight of Solute
.$$ \times $$ 10
= $${{63 \times 1.4 \times 10} \over {63}}$$ = 14 M
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