B) The student then does the titration correctly. He finds that 16.70 cm3 of the dilute sulfuric acid neutralises 25.0 cm3 of sodium hydroxide solution of concentration 0.200 mol/dm3

The equation for the reaction is

2NaOH + H2SO4 → Na2SO4 + 2H2O

ii) Calculate the amount, in moles, of sulfuric acid

Explanation:

mole of acid/mole of base=(concentration of acid×volume of acid)/(concentration of base×volume of base)

Na/Nb=CaVa/CbVb

1/2=(Ca×16.7)/(0.2×25)

0.2×25=2×Ca×16.7

Ca=(0.2×25)/(2×16.7)

Ca=0.15mol/dm³

mole of acid=Concentration×volume

mole of acid=0.15×(16.7/1000)

mole of acid=2.5×10^-3moles

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.