A piece of zinc weighing 0.635 g when treated with excess of dilute H2SO4 liberated 200 cm^3 hydrogen at STP.

Given : 

Mass of zinc = 0.635 g, 

Volume of H₂ liberated = 200 cm³

To find : 

% purity of zinc sample

Calculation : 

The relevant balanced chemical equation is,

Zn + H₂SO₄ → ZnSO₄ + H₂

It indicates that 22.4 L of hydrogen at STP = 65 g of Zn.

(where, atomic mass of Zn = 65 u)

∴ 0.200 L of hydrogen at STP

= \(\frac{65g}{22.4L}\) x 200 L

= 0.5803 g of Zn

∴ Percentage purity of Zn = \(\frac{0.5803}{0.635}\) × 100

= 91.37 % (by using log tables)

∴ Percentage purity of Zn sample = 91.37%

[Calculation using log table : \(\frac{65\times 0.200}{22.4}\) 

= Antilog_10[log₁₀(65) + log₁₀(0.200) - log₁₀(22.4)]

= Antilog₁₀[1.8129 + \(\bar 1.3010\) – 1.3502]

= Antilog₁₀\(\overline{1} .7637\)

= \(\frac{0.5803}{0.635}\) × 100

= \(\frac{58.03}{0.635}\) 

= Antilog₁₀[log₁₀(58.03) – log₁₀(0.635)]

= Antilog₁₀[1.7636 – \(\bar 1.8028\)]

= Antilog₁₀[1.9608] 

= 91.37