Answer: The mass percent of water in [tex]Na_2CO_3.10H_2O[/tex] is 63%.
Explanation:
In [tex]Na_2CO_3.10H_2O[/tex], there are 2 sodium atoms, 1 carbon atom 13 oxygen atoms and 20 hydrogen atoms.
To calculate the mass percent of element in a given compound, we use the formula:
[tex]\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Molar mass of}Na_2CO_3.10H_2O}\times 100[/tex]
Mass of water = [tex]10\times 18g/mol=180g[/tex]
Mass of [tex]Na_2CO_3.10H_2O[/tex]= 286 g
Putting values in above equation, we get:
[tex]\text{Mass percent of water}=\frac{180}{286}\times 100=63\%[/tex]
Hence, the mass percent of water in [tex]Na_2CO_3.10H_2O[/tex] is 63%
