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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» Solution :Sodium Nitrate = `NaNO_(3)` Molecular MASS of Sodium Nitrate = NaNO3 = 23 + 14 + 48 = 85 100% pure 85 g of `NaNO_(3)` CONTAINS 23 g of Sodium. 100% pure `95xx10^(3)` g of `NaNO_(3)` will contains` 23/85 xx95 xx 10^(3)` = `25.70xx10^(3)` g of Sodium. 100% pure NaNO3 contains `25.70xx10^(3)` g of Sodium. `:. `100% pure `NaNO_(3)` contains `25.70xx10^(3)`g of sodium `:. `70% pure `NaNO_(3)` will contains `(25.70xx10^(3))/100 xx 70` 17999 (g) or 17.99 kg of Na |
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