Answer
See answer below.
Work Step by Step
a)Average volume: $(20.15+21.30+20.40+20.35)/4=20.55\ mL$
Number of moles of NaOH:
$20.55\ mL\times 0.1760\ M=3.617\ mmol$
Number of moles of sulfuric acid: $1.808\ mmol$
Concentration: $1.808\ mmol/10.00\ mL=0.181\ M$
Concentration before dilution:
$0.181\ M\times 500.\ mL\div 5.00\ mL=18.08\ M$
b) Mass: $18.08\ M\times 98.07\ g/mol=1.773\ kg$
Mass percentage:
$1.773/1.84\times100\%=96.38$
