JEE MAIN - Chemistry (2020 - 3rd September Evening Slot - No. 18)
6.023 $$ \times $$ 1022 molecules are present in 10 g of a substance 'x'. The molarity of a solution containing
5 g of substance 'x' in 2 L solution is _____ × 10-3
Answer
25
Explanation
Mass of 6.023 × 1022 molecules of a substance
= 10 g
Mass of 6.023 × 1023 molecules of the substance = 100 g
$$ \therefore $$ Molar mass of the substance = 100 g mol–1
Molarity of the solution =
= $${{\left( {5/100} \right)} \over 2}$$ = 0.025
Mass of 6.023 × 1023 molecules of the substance = 100 g
$$ \therefore $$ Molar mass of the substance = 100 g mol–1
Molarity of the solution =
moles of solute
volume of solution(in l)
= $${{\left( {5/100} \right)} \over 2}$$ = 0.025
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