Determine the mass in grams of 0.5 moles of sulphuric acid this mass was disolved in 1 dm3 of distilled water?

calculate the volume of the sulphuric acid that would neutralise 25cm3 of sodium hydroxide solution with a concentration of 0.2 moldm3

Answers:
step 1 2NaOH + H2SO4 ---------> Na2SO4 + 2H2O
step 2 amount of NaOH = 0.2 x 25/1000= 5.0 x 10^-3 mol
step 3 amount of H2SO4 = s.0x10^-3 x 1/2 = 2.5 x 10^-3 mol
step 4 mass of H2SO4 = 2.5 x 10^-3 x 98 = 0.245g required

NOTE
^= power
in step 4 it is multiplied by 98 because that is the molar mass of H2SO4 ie (2x1)+32+ (16 x 4) =98

i do hope this is correct it has a been about 3 yrs since i did my A Level in chemistry
First part.
The solution you describe is a 0.5 molar solution.
The relative molecular mass of sulphuric acid is 98 so 98g would make a 1 Molar solution.
A 0.5 molar solution therefore contains 0.5 x 98g = 49g
2nd Part
From the equation for the neutalisation:

H2SO4 + 2NaOH = Na2SO4 + 2H2O

You can see that 2 moles of NaOH are needed to neutalise 1 mole of H2SO4. (half as many moles of acid as alkali)

1 litre of 1 M NaOH needs 0.5 litres 1 M sulphuric acid (thats 0.5 M)

So 1 litre of 1 M NaOH needs 1 litre of 0.5M sulphuric acid (still 0.5 Mole)

and 25 cm3 of 1M NaOH needs 25cm3 of 0.5M sulphuric acid

therefore 25cm3 of 0.2 M NaOH need 25 x 0.2 cm3 of 0.5 M acid

= 5cm3

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