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