JEE MAIN - Chemistry (2021 - 17th March Morning Shift - No. 20)
The mole fraction of a solute in a 100 molal aqueous solution is ___________ $$\times$$ 10$$-$$2. (Round off to the Nearest Integer).
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
Answer
64
Explanation
Let weight of H2O = 1000 g
Moles of solute = 100
(mole)H2O = $${{1000} \over {18}}$$
Mole fraction of solute = $${{mole\,of\,solute} \over {Total\,moles}}$$
$$ = {{100} \over {100 + {{1000} \over {18}}}} = {{1800} \over {2800}}$$
$${X_{solute}} = 64 \times {10^{ - 2}}$$
Moles of solute = 100
(mole)H2O = $${{1000} \over {18}}$$
Mole fraction of solute = $${{mole\,of\,solute} \over {Total\,moles}}$$
$$ = {{100} \over {100 + {{1000} \over {18}}}} = {{1800} \over {2800}}$$
$${X_{solute}} = 64 \times {10^{ - 2}}$$
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