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