2.5 g of a hydrate barium salt gave on heating, 2.13 g of the anhydrous salt. Given that the relative molecular mass of the of the anhydrous salt is 208, the number of molecules of water of crystallization of the barium salt is
Given:
Mass of hydrated Barium salt = 2.5g
Mass of anhydrous Barium salt = 2.13g
Molar mass of anhydrous salt = 208
Molar mass of anhydrous salt = 208 + 18x ( 18x due to the unknown molecule of water of crystallization, i.e xH2O, since Molar mass of H2O = 18, then multiplied by X)
To calculate the number of water of crystallization, we use the formula:
\(\frac{mass of hydrated salt}{mass of anhydrous salt}\) = \(\frac{Molar mass of hydrated salt}{Molar mass of anhydrous salt}\)
\(\frac{2.5}{2.13}\) = \(\frac{208 + 18x}{208}\)
2.13 (208 + 18x) = 2.5 x 208
443.04 + 38.34x = 520
38.34x = 520 - 443.04
38.34x = 76.96
x = \(\frac{76.96}{38.34}\)
x = 2.00
Therefore, the number of water of crystallization is 2
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