Answer:
Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.
Explanation:
From the given,
Mass of sodium carbonate [tex]Na_{2}CO_{3}.XH_{2}O[/tex] = 8.85 g
Loss mass [tex]H_{2}O[/tex] = 1.28 g
Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g
[tex]7.57 g Na_{2}CO_{3} \times \frac{1mol}{106 g} =\frac{0.0714}{0.0714}=1[/tex]
[tex]1.28g H_{2}O \times \frac{1mol}{18 g} =\frac{0.0711}{0.0714}=1[/tex]
Therefore, the compound has only one water molecule.
Molecular formula of the compound is [tex]Na_{2}CO_{3}.H_{2}O[/tex] an name of the compound is sodium carbonate mono hydrate.
Hence, the type of the compound is Mono hydrate.
