JEE MAIN - Chemistry (2019 - 9th April Evening Slot - No. 9)
What would be the molality of 20% (mass/
mass) aqueous solution of KI?
(molar mass of KI = 166 g mol–1)
(molar mass of KI = 166 g mol–1)
1.51
1.35
1.08
1.48
Explanation
20% (mass/
mass) aqueous solution means,
100 g solution contains 20 g KI
$$ \therefore $$ Mass of solvent = 100 – 20 = 80 g
So, 20 g of KI present in 80 g of solvent.
$$ \therefore $$ $${{20} \over {166}}$$ moles of KI present in 80 g of solvent.
We know,
Molality (m) = $${{no\,of\,moles\,of\,solute} \over {wt(in\,kg)\,of\,solvent}}$$
$$ \therefore $$ m = $${{{{20} \over {166}}} \over {{{80} \over {1000}}}}$$ = 1.51
100 g solution contains 20 g KI
$$ \therefore $$ Mass of solvent = 100 – 20 = 80 g
So, 20 g of KI present in 80 g of solvent.
$$ \therefore $$ $${{20} \over {166}}$$ moles of KI present in 80 g of solvent.
We know,
Molality (m) = $${{no\,of\,moles\,of\,solute} \over {wt(in\,kg)\,of\,solvent}}$$
$$ \therefore $$ m = $${{{{20} \over {166}}} \over {{{80} \over {1000}}}}$$ = 1.51
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