Question

Describe how you can prepare the following solution. A 492.6 g solution of a 1.98 % (w/w) benzene solution in chloroform. Calculate the morality of an aqueous solution made with 768.93 mg of (NH_4)_2SO_4 to a final volume of 675 mL?

Answer 1

Solution:

A. Given Data:

492.6 g of solution are preparing with concentration of 1.98 % w/w benzene in chloroform

We have total solution of 492.6 gram.

Quantity of benzene in this solution = 492.6 * 0.0198 = 9.753 gram benzene

Quantity of Chloroform = 492.6 - 9.753 = 482.84 gram of chloroform.

So to prepare this perticular solution weight 9.753 gram of benzene and 482.84 gram of chloroform and mix them. The resulting solution will be 1.98 % of Benzene in chloroform.

B. Given Data:

mass of solute = 768.93 gram of (NH4)2SO4

moles of solute = mass/Molecular weight

= 768.93/132

= 5.825 moles

Volume of Solution = 675 mL = 0.675 L

Molarity = Moles of Solute/Vol of Solution in liter

= 5.825/0.675

Molarity = 8.63 M