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On heating 1.763 g of hydrated $B a C l_{2} . n H_{2} O$ to dryness, 1.505 g of anhydrous salt remained. What is the value of ‘n’ (Mol. wt. of $B a C l_{2} = 208$ )

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Detailed Solution

$\begin{array}{l} \frac{208 + 18 n}{1.763} = \frac{208}{1.505} \\ ∴ n = 1.98 ≃ 2 \end{array}$

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