Question

A hydrate of $Na_2SO_3$ has 50% water by mass. It is:

• A. $Na_2SO_3 \cdot 5H_2O$
• B. $Na_2SO_3 \cdot 6H_2O$
• C. $Na_2SO_3 \cdot 7H_2O$
• D. $Na_2SO_3 \cdot 2H_2O$

Answer 1

Answer: (C)

$Na_2SO_3 \cdot 7H_2O$

Answer 2

1. Calculate the molar mass of anhydrous sodium sulfite ($Na_2SO_3$) and water ($H_2O$).
   • Molar mass of $Na_2SO_3 = (2 \times 23) + 32 + (3 \times 16) = 46 + 32 + 48 = 126$ g/mol.
   • Molar mass of $H_2O = (2 \times 1) + 16 = 18$ g/mol.
2. The hydrate contains 50% water by mass, which implies the anhydrous salt ($Na_2SO_3$) also constitutes 50% of the total mass. Therefore, the mass of $Na_2SO_3$ is equal to the mass of water in the hydrate.
3. To find the number of water molecules ($x$) in the hydrate formula $Na_2SO_3 \cdot xH_2O$, we determine the ratio of moles of water to moles of $Na_2SO_3$. Since their masses are equal, this ratio is: $x = \frac{\text{Molar mass of } Na_2SO_3}{\text{Molar mass of } H_2O} = \frac{126 \text{ g/mol}}{18 \text{ g/mol}} = 7$.
4. The formula of the hydrate is $Na_2SO_3 \cdot 7H_2O$.