Answer :
On balancing the neutralization reaction of sulphuric acid and sodium hydroxide. From the stoichiometry of balanced equation, we can say that for every 2 moles of NaOH we need one mole of sulfuric acid.
From given data we know that molarity of NaOH is 0.187M and the volume is 20 ml
Also, molarity= no of moles / Volume
No. of moles= molarity × volume
No of moles of NaOH = 0.187 × 20/1000 = 3.74 × 10(-3) moles
Hence no of moles of sulphuric acid will be half of the number of moles of NaOH i.e. 1.87 × 10(-3) moles.
Again we know that
molarity= no of moles / Volume
So, the volume= no of moles/ molarity
Volume of sulphuric acid= 1.87 × 10(-3)/0.500
= 3.74 × 10(-3) litres
= 3.74 ml.
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