Respuesta :
We can calculate this in the following way;
%MgSO4 in MgSO4.7H2O = [(molar mass MgSO4/molar mass MgSO4.7H2O)] x 100 = ?
Molar mass of MgSO4.7H2O is 246.4746 g/mol
Molar mass of MgSO4 is 120.366 g/mol.
%MgSO4 in MgSO4.7H2O = 120.366 / 246.4746 = 0.488 x 100 = 48.8
Then 2.4g x (%MgSO4/100) = 2.4 x (48.8/100) = 2.4 x 0.488 = 1.17 = 1.2g MgSO4
%MgSO4 in MgSO4.7H2O = [(molar mass MgSO4/molar mass MgSO4.7H2O)] x 100 = ?
Molar mass of MgSO4.7H2O is 246.4746 g/mol
Molar mass of MgSO4 is 120.366 g/mol.
%MgSO4 in MgSO4.7H2O = 120.366 / 246.4746 = 0.488 x 100 = 48.8
Then 2.4g x (%MgSO4/100) = 2.4 x (48.8/100) = 2.4 x 0.488 = 1.17 = 1.2g MgSO4
The quantity of pure [tex]\rm \bold {MgSO_4}[/tex] in 2.4 g of [tex]\rm \bold {MgSO_4.7H_2O}[/tex] is 1.17 g.
Molar mass of [tex]\rm \bold {MgSO_4.7H_2O}[/tex] is 246.4746 g/mol
Molar mass of [tex]\rm \bold {MgSO_4}[/tex] is 120.366 g/mol.
Means for 246.4746 g of [tex]\rm \bold {MgSO_4.7H_2O}[/tex], The amount of [tex]\rm \bold {MgSO_4}[/tex] is 120.366 g.
Hence,
For 2.4 g of [tex]\rm \bold {MgSO_4.7H_2O}[/tex] The amount of [tex]\rm \bold {MgSO_4}[/tex] will be
[tex]\rm \bold {\Rightarrow 2.4 \times \frac{120.366 }{246.47} } \\\\\rm \bold {\Rightarrow 2.4\times 0.48}\\\rm \bold {\Rightarrow 1.17}[/tex]
Hence, we can conclude that the quantity of pure [tex]\rm \bold {MgSO_4}[/tex] in 2.4 g of [tex]\rm \bold {MgSO_4.7H_2O}[/tex] is 1.17 g.
To know more about molar mass you can refer to the link:
https://brainly.com/question/22997914?referrer=searchResults
