volume of 18M H2SO, taken for preparing 0.5M H2SO4(100 ml solution)
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This is like dilution.
We take an acid of high molarity then we dilute it to get low molarity.
The formula that helps in solving this problem is:
m1v1 = m2v2
Where:
m1 = initial molarity = 18M
v1 = this is the initial volume (it is what is required)
m2 = final molarity = 0.5M
v2 = final volume = 100ml = 100/1000 = 0.1 litre
therefore;
18v1 = 0.5x0.1
v1 = (0.5x0.1)/18
= 0.00278 litres
= 2.78 ml
So, in preparing 100ml of 0.5M acid solution, we take 2.78ml of the 18M acid
We take an acid of high molarity then we dilute it to get low molarity.
The formula that helps in solving this problem is:
m1v1 = m2v2
Where:
m1 = initial molarity = 18M
v1 = this is the initial volume (it is what is required)
m2 = final molarity = 0.5M
v2 = final volume = 100ml = 100/1000 = 0.1 litre
therefore;
18v1 = 0.5x0.1
v1 = (0.5x0.1)/18
= 0.00278 litres
= 2.78 ml
So, in preparing 100ml of 0.5M acid solution, we take 2.78ml of the 18M acid
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