Complete Question
Magnesium sulfate forms a hydrate with the formula [tex]MgSO_4. 7H_20[/tex]. What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of [tex]MgSO_4[/tex] is 120.4 g/mol; H20 = 18 g/mol.
Answer:
The value is [tex]z = 0.2093 \ g[/tex] of [tex]H_2O[/tex]
Explanation:
From the question we are told that
The volume of toluene is [tex]V = 15 mL[/tex]
The mass of anhydrous magnesium sulfate is [tex]m = 200m g = 200 *10^{-3} \ g[/tex]
The formula of the hydrate is [tex]MgSO_4. 7H_20[/tex]
The molar mass of [tex]MgSO_4[/tex] is [tex]z =120.4 \ g/mol[/tex]
From the formula given we see that
1 mole of [tex]Mg SO_4[/tex] wil remove 7 moles of [tex]H_2O[/tex] to for the given formula
Hence
120.4 g (1 mole) will remove 7 moles (7 * 18 g = 126 g ) of [tex]H_2O[/tex] to for the given formula
Therefore 1 g of [tex]Mg SO_4[/tex] x g of [tex]H_2O[/tex]
So
[tex]x = \frac{x]126 * 1}{ 120.4 }[/tex]
=> [tex]x = 1.0465 \ g [/tex]
From our calculation we obtained that
1 g of [tex]Mg SO_4[/tex] will remove [tex]x = 1.0465 \ g [/tex] of [tex]H_2O[/tex]
Then
[tex]200 *10^{-3} \ g[/tex] of [tex]Mg SO_4[/tex] will remove z g of [tex]x = 1.0465 \ g [/tex] of [tex]H_2O[/tex]
So
[tex]z = 200 *10^{-3} * 1.0465[/tex]
=>[tex]z = 200 *10^{-3} * 1.0465[/tex]
=>[tex]z = 0.2093 \ g[/tex] of [tex]H_2O[/tex]
