1.

Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.

Answer»

Sodium Nitrate = NaNO3 

Molecular mass of Sodium Nitrate = 23 + 14 + 48 = 85 

100% pure 85 g of NaNO3 contains 23 g of Sodium. 

100% pure 95 x 103 g of NaNO3 will contains 23/85 x 95 x 103 

= 25.70 x 103 g of Sodium.

100% pure NaNO3 contains 25.70 x 103 g of Sodium. 

∴ 70% pure NaNO3 will contains = 1 mg 

= 17990 g (or) 17.99 Kg of Na.



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