Chemical Engineering - Stoichiometry - Discussion

Discussion Forum : Stoichiometry - Section 1 (Q.No. 49)
49.
The weight fraction of methanol in an aqueous solution is 0.64. The mole fraction of methanol XM satisfies
XM < 0.5
XM = 0.5
0.5 < XM < 0.64
XM ≥ 0.64
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
6 comments Page 1 of 1.

TANMAY SHAH said:   2 years ago
Aqueous solution of Methanol means a solution containing Methanol + Water.

Methanol CH3OH = 32 M.W.
Water H2O = 18 M.W.

Now, Assume 1 g of total solution.
It contains 0.64 g of Methanol &
It contains 0.36 g of Water.

Now, Moles = Mass/M.W.

Moles of Methanol = 0.64/32 = 0.02 moles
Moles of Water = 0.36/18 = 0.02 moles.

Total Moles = 0.02 + 0.02 = 0.04 moles.

Now Mole fraction of Methanol = 0.02/0.04 = 0.5.

So, Option B is the Right Answer.
(1)

Abdullah said:   6 years ago
(0.64g CH3OH/ 1g solution)(1mol CH3OH/ 32g CH3OH)(1g solution/ 0.04mol solution) = 0.5mol CH3OH/ mol solution.

So, I think, the answer is 0.5.

Rajiv Kumar Balayan said:   7 years ago
I think B is correct.

Sayyad said:   7 years ago
Mol fraction = vol fra = weight fraction.

Vishal said:   7 years ago
Anyone explain it.

John said:   10 years ago
Answer B in 100 kg solution Mol of methanol = 64/32 = 2.

Mol of water = 36/18 = 2 so, xm = 2/4 = 0.5.
(1)

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