In an experiment to determine Avogadro's number, a charge of $965 C$ passed through aqueous $CuSO X_{4}$ deposits $0.3175 g$ of copper at the cathode. Taking the electronic charge as $1.60 \times 1 0^{- 19} C$ and Cu = 63.5, find the value of Avogadro's number obtained, expressed as $N_{A} \times 1 0^{- 23}$ (two decimals).

Numerical answer type, tolerance 0.02.

Answer: 6.03

Number of electrons passed

= 965/ (1.60 \times 1 0^{- 19}) = 6.031 \times 1 0^{21}

. Moles of Cu

= 0.3175/63.5 = 5.00 \times 1 0^{- 3} mol

; since

Cu X^{2 +} + 2 e X^{-} Cu

, moles of electrons

= 1.00 \times 1 0^{- 2}

. Hence

N_{A} = 6.031 \times 1 0^{21} /1.00 \times 1 0^{- 2} = 6.03 \times 1 0^{23} mo l^{- 1}

, so the answer is

6.03