Respuesta :
Answer:
[tex]NaSO_4.10H_2O[/tex]
Explanation:
Given that:-
Mass of the hydrated salt = 7.028 g
Mass of the anhydrous salt = 3.100 g
Mass of water eliminated = Mass of the hydrated salt - Mass of the anhydrous salt = 7.028 - 3.100 g = 3.928 g
Moles of water:
Mass of water = 3.928 g
Molar mass of [tex]H_2O[/tex] = 18 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus, moles are:
[tex]moles= \frac{3.928\ g}{18\ g/mol}[/tex]
[tex]moles_{water}= 0.2212\ mol[/tex]
Moles of anhydrous salt:
Amount = 3.100 g
Molar mass of [tex]NaSO_4[/tex] = 142.04 g/mol
Thus, moles are:
[tex]moles= \frac{3.100\ g}{142.04\ g/mol}[/tex]
[tex]moles_{CaSO_4}= 0.02182\ mol[/tex]
The simplest ratio of the two are:
[tex]NaSO_4:H_2O[/tex] =0.02182 : 0.2212 = 1 : 10
Hence, the formula for hydrate is:- [tex]NaSO_4.10H_2O[/tex]
