JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 12)
8 g of NaOH is dissolved in 18g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are -
0.2, 11.11
0.167, 22.20
0.167, 11.11
0.2, 22.20
Explanation
8 gm of NaOH = $${8 \over {40}}$$ = 0.2 mol of NaOH
18 gm of H2O = $${18 \over {18}}$$ = 1 mol of H2O
$$ \therefore $$ Total mole = 1 + 0.2 = 1.2 mol
$$ \therefore $$ Mole fraction of NaOH = $${{0.2} \over {1.2}}$$ = 0.167
We know,
Molality = $${{Moles\,\,of\,\,solute} \over {Weight\,of\,solvent}} \times $$ 1000
= $${{0.2} \over {18}} \times $$ 1000
= 11.11
18 gm of H2O = $${18 \over {18}}$$ = 1 mol of H2O
$$ \therefore $$ Total mole = 1 + 0.2 = 1.2 mol
$$ \therefore $$ Mole fraction of NaOH = $${{0.2} \over {1.2}}$$ = 0.167
We know,
Molality = $${{Moles\,\,of\,\,solute} \over {Weight\,of\,solvent}} \times $$ 1000
= $${{0.2} \over {18}} \times $$ 1000
= 11.11
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