JEE MAIN - Chemistry (2019 - 12th April Morning Slot - No. 14)
The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg–1
) of the
aqueous solution is :
13.88 × 10–1
13.88 × 10–3
13.88
13.88 × 10–2
Explanation
We know,
molality(m) = $${{{x_{solute}}} \over {{x_{solvent}}}} \times {{1000} \over {molar\,weight\,of\,solvent}}$$
Here solvent is H2O
xsolvent = 0.8
$$ \therefore $$ xsolute = 0.2
$$ \therefore $$ Molality = $${{0.2} \over {0.8}} \times {{1000} \over {18}}$$ = 13.88
molality(m) = $${{{x_{solute}}} \over {{x_{solvent}}}} \times {{1000} \over {molar\,weight\,of\,solvent}}$$
Here solvent is H2O
xsolvent = 0.8
$$ \therefore $$ xsolute = 0.2
$$ \therefore $$ Molality = $${{0.2} \over {0.8}} \times {{1000} \over {18}}$$ = 13.88
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