Answer: 127.5ml
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
Thus 127.5 ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH
C
i think i am sorry if am wrong
The approximate alcohol content is 210 ml.
Explanation:
It can be deduced from the question that each bottle is of 1000ml or 1 litre.
The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is
20/100*500
=100 ml
The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml
so it is 200ml having 30% alcohol
30/100*200
= 60 ml
The third bottle is one tenth full so its volume is 1/10*1000
100 ml. having 50% of alcohol
50/100*100
50 ml.
The alcohol content obtained from all these 3 litres is:
100+60+50
= 210 ml of alchohol is obtained from 800 ml of mixture.
