Question

What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?

Answer 1

Complete Question

Magnesium sulfate forms a hydrate with the formula $MgSO_4. 7H_20$. 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 $MgSO_4$ is 120.4 g/mol; H20 = 18 g/mol.

Answer:

The value  is  $z = 0.2093 \ g$ of  $H_2O$

Explanation:

From the question we are told that

   The volume of toluene is  $V = 15 mL$

    The mass of  anhydrous magnesium sulfate is  $m = 200m g = 200 *10^{-3} \ g$

   The formula of the hydrate is   $MgSO_4. 7H_20$

    The molar mass of   $MgSO_4$  is  $z =120.4 \ g/mol$

From the formula given we see that

  1 mole of  $Mg SO_4$ wil remove  7 moles of $H_2O$ to for the given formula

Hence

  120.4 g (1 mole) will remove  7 moles (7 * 18 g = 126 g  ) of  $H_2O$ to for the given formula

Therefore 1 g of  $Mg SO_4$  x g  of  $H_2O$  

So

     $x = \frac{x]126 * 1}{ 120.4 }$

=>     $x = 1.0465 \ g$

From our calculation we obtained that

  1 g of $Mg SO_4$ will remove  $x = 1.0465 \ g$  of  $H_2O$  

Then  

   $200 *10^{-3} \ g$ of $Mg SO_4$ will remove z g of  $x = 1.0465 \ g$  of  $H_2O$  

So

   $z = 200 *10^{-3} * 1.0465$

=>$z = 200 *10^{-3} * 1.0465$

=>$z = 0.2093 \ g$ of  $H_2O$