A vessel is full of a mixture of methanol and ethanol in which there ...
Problem:
A vessel is full of a mixture of methanol and ethanol in which there is 20% ethanol. 10 litres of mixture are drawn off and filled with methanol. If the ethanol is now 15%, what is the capacity of the vessel?
Solution:
Let us assume that the vessel has a capacity of x liters.
Initial Mixture:
Initially, the mixture in the vessel contains 20% ethanol.
So, the amount of ethanol in the initial mixture = (20/100) * x = 0.2x liters.
And, the amount of methanol in the initial mixture = (80/100) * x = 0.8x liters.
When 10 liters of the initial mixture is drawn off, the amount of ethanol and methanol in the mixture will be:
Ethanol: (20/100) * 10 = 2 liters
Methanol: (80/100) * 10 = 8 liters
After drawing off 10 liters of the mixture, the vessel contains (x - 10) liters of the initial mixture.
Now, 10 liters of methanol is added to the mixture. So, the amount of methanol in the mixture will be:
Methanol: 8 + 10 = 18 liters
And, the amount of ethanol in the mixture will remain the same, i.e., 2 liters.
Final Mixture:
The final mixture contains 15% ethanol.
So, the amount of ethanol in the final mixture = (15/100) * x liters.
And, the amount of methanol in the final mixture = (85/100) * x liters.
We know that the amount of ethanol in the final mixture is (amount of ethanol in the initial mixture - 10 + 0) liters.
So, we can write the equation:
(15/100) * x = 0.2x - 10
Solving this equation, we get:
x = 40 liters
Therefore, the capacity of the vessel is 40 liters.
Answer: (a) 40 liters
A vessel is full of a mixture of methanol and ethanol in which there ...
Therefore, Mixture: Methanol = 3:1
So, for 10 litres of methanol, 30 I of Mixture is needed. So, total capacity = (30 + 10)l = 40l