A) 50 mol per cent
B) 54 mol per cent
C) 32 mol per cent
D) 44 mol percent
Correct Answer: A
Solution :
Let the amount of 'A' in the mixture be \[{{\chi }_{A}}\] and B be \[{{\chi }_{B}}\] \[{{p}_{T}}=760={{p}^{\circ }}_{A}{{\chi }_{A}}+{{p}^{\circ }}_{B}{{\beta }_{B}}\] \[760=520{{\chi }_{A}}+1000(1-{{\chi }_{A}})\] \[480{{\chi }_{A}}=240\] \[{{\text{ }\!\!\chi\!\!\text{ }}_{\text{A}}}\text{=}\frac{\text{1}}{\text{2}}\text{or 50 mol}\] percentYou need to login to perform this action.
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