How many moles of oxygen will occupy a volume of 2.50 at 1.20 atm and 25.0 c?
1 answer:
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Answer:
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
Explanation:
Let the mass of the first solution be x and second solution be y.
Amount solution required = 1250 kg
x + y = 1250 kg....[1]
Percentage of ethanol in required solution = 12% of 1250 kg
Percentage of ethanol in solution-1 = 5% of x
Percentage of ethanol in required solution = 25% of y
5% of x + 25% of y =12% of 1250 kg
x + 5y = 3000 kg...[2]
Solving [1] and [2] we :
x = 437.5 kg , y = 812.5 kg
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.