When a crystalline salt Na_2SO_4 cdot xH_2O i | vedclass

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Question

(D) Let the initial weight of the hydrated salt be $100 \ g$.Weight of water lost $= 55.9 \ g$.Weight of anhydrous $Na_2SO_4$ remaining $= 100 - 55.9

Vedclass · Accepted Answer

$Na_2SO_4 \cdot 10H_2O$

Vedclass · Answer

(D) Let the initial weight of the hydrated salt be $100 \ g$.Weight of water lost $= 55.9 \ g$.Weight of anhydrous $Na_2SO_4$ remaining $= 100 - 55.9 = 44.1 \ g$.Molar mass of anhydrous $Na_2SO_4 = (2 	imes 23) + 32 + (4 	imes 16) = 142 \ g/mol$.$44.1 \ g$ of $Na_2SO_4$ is associated with $55.9 \ g$ of $H_2O$.Therefore,$142 \ g$ of $Na_2SO_4$ is associated with $\frac{142 	imes 55.9}{44.1} \approx 180 \ g$ of $H_2O$.Number of moles of water $(x)$ $= \frac{180}{18} = 10$.Thus,the formula of the crystalline salt is $Na_2SO_4 \cdot 10H_2O$.

When a crystalline salt $Na_2SO_4 \cdot xH_2O$ is heated,it loses $55.9\%$ of its weight. What is the formula of the crystalline salt?

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