(a)

Burette reading/Titration	1st Titration	2nd Titration	3rd Titration
Final burette reading (cm\(^3\))	16.50	16.10	21.10
Initial burette reading (cm\(^3\))	0.00	0.00	5.00
Average volume of acid used	16.50	16.10	16.10

The first titration as a trial
Average volume of acid used
= \(\frac{16.10 + 16.10cm^3}{2}\)
= \(\frac{32.20cm^3}{2}\)
= 16.10cm\(^3\)

(b)(i) Concentration of A in mol/dm\(^3\)
500cm\(^3\) of solution contains 5.00g of HNO\(_3\)
100cm\(^3\) of solution will contain
\(\frac{5.00}{500} \times \frac{100}{1}\) = 10.00 gdm\(^{-3}\)
Molar mass of HNO\(_3\) = (1 x 1) + (14 x 1) + (16 x 3)
= 1 + 14 + 48 = 63gmol\(^{-1}\)
Using the relation:
Concentration g/dm\(^3\)
= molar conc.moldm\(^{-3}\) x molar mass
10.00 = molar conc. moldm\(^{-3}\) x 63
Concentration of A = \(\frac{10.00}{63}\)
= 0.15873015873 moldm\(^{-3}\)
= 0.158 moldm\(^{-3}\)


(ii) Concentration of B in moldm\(^{-3}\)
Using the formula
\(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{0.159 \times 16.10}{C_B \times 25.0} = \frac{1}{1}\)
\(C_B = \frac{0.159 \times 16.10 \times 1}{25.0 \times 1}\)
= 0.0102396 moldm\(^{-3}\)
\(C_B\) = 0.102 moldm\(^{-3}\)

(iii) Concentration of B in gdm\(^{-3}\)
concentration g/dm\(^{-3}\) = molar concentration of moldm\(^{-3}\) x molar mass
Molar mass NaOH = 40g/mol
Concentration of B = 0.102 x 40
= 4.08g/dm\(^3\)

(iv) Mass of NaOH
No. of moles of NaOH = \(\frac{\text{molar conc.moldm}^{-3} \times volume}{100}\)
= \(\frac{0.102 \times 250}{1000}\) = 0.0255 mole
from the equation
1 mole of NaOH = 0.055mole of NaNo\(_3\)
0.0255 mole of NaOH = 0.0255 mole of NaNO\(_3\)
molar mass of NaNO\(_3\) = 85 gmol\(^{-1}\)
mass of NaNO\(_3\) = 0.0255 x 85
= 2.1675g