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.
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Hope this helps :)
Answer:
0.2 M.
Explanation:
- For the acid-base neutralization, we have the role:
The no. of millimoles of acid is equal to that of the base at the neutralization.
<em>∴ (XMV) KOH = (XMV) H₂SO₄.</em>
X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),
M is the molarity.
V is the volume.
X = 1, M = 0.5 M, V = 38.74 mL.
X = 2, M = ??? M, V = 50.0 mL.
∴ M of H₂SO₄ = (XMV) KOH/(XV) H₂SO₄ = (1)(0.5 M)(38.74 mL)/(2)(50.0 mL) = 0.1937 M ≅ 0.2 M.