The hydrated salt $N a_{2} C O_{3} . x H_{2} O$ undergoes $63 %$ loss in mass on heating and becomes anhydrous. The value of $x$ is: (report the answer to the closest integer)

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Solution

$N a_{2} C O_{3} . x H_{2} O (s) \to N a_{2} C O_{3} (s) + x H_{2} O (g)$
Let $100 g$ of $N a_{2} C O_{3} . x H_{2} O$ be present.
$∴ \frac{100 x}{106 + 18 x} mole of H_{2} O formed$
$63 = \frac{100 x}{106 + 18 x} \times 18$
$6678 + 1134 x = 1800 x$
$666 x = 6678$
$x = 10$

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The hydrated salt Na2CO3.xH2O undergoes 63% loss in mass on heating and becomes anhydrous. The value of x is: (report the answer to the closest integer)

Na2CO3.xH2O(s)→Na2CO3(s)+xH2O(g) Let 100 g of Na2CO3.xH2O be present. ∴100x106+18x mole of H2O formed 63=100x106+18x×18 6678+1134x=1800x 666x=6678 x=10

The hydrated salt $N a_{2} C O_{3} . x H_{2} O$ undergoes $63 %$ loss in mass on heating and becomes anhydrous. The value of $x$ is: (report the answer to the closest integer)

$N a_{2} C O_{3} . x H_{2} O (s) \to N a_{2} C O_{3} (s) + x H_{2} O (g)$
Let $100 g$ of $N a_{2} C O_{3} . x H_{2} O$ be present.
$∴ \frac{100 x}{106 + 18 x} mole of H_{2} O formed$
$63 = \frac{100 x}{106 + 18 x} \times 18$
$6678 + 1134 x = 1800 x$
$666 x = 6678$
$x = 10$