Question

1. Describe how you would prepare each of the following aqueous solutions.

     a. 1.30 L of 0.130 M (NH4)2SO4 solution, starting with solid (NH4)2SO4.   Enter your answers numerically separated by a comma.

     b. 120 g of a solution that is 0.55 m in Na2CO3, starting with the solid solute, Enter your answers numerically separated by a comma.

     c. 1.30 L of a solution that is 15.0 % of Pb(NO3)2 by mass (the density of the solution is 1.16 g/mL), starting with solid solute, Enter your answers numerically separated by a comma.

     d. a 0.50M solution of HCl that would just neutralize 6.5 g of Ba(OH)2 starting with 6.0 M HCl, Enter your answers numerically separated by a comma.

Answer 1

a) Moles of (NH4)2SO4 in solution = 1.3 x 0.13 = 0.169 moles

Molar Mass of (NH4)2SO4 = 132.14 g / mol

=> Mass of (NH4)2SO4 required = 0.169 x 132.14 = 22.33 g

Therefore, we add 22.33 g of (NH4)2SO4 to make the solution

b)

Molality (m) = Moles of solute / Mass of solvent (in kg)

Let mass of Na2CO3 added = m gram

=> Moles of Na2CO3 = m / 106 moles

=> Mass of solvent = 120 - m

=> 0.55 = (m / 106) / ((120-m) / 1000)

=> 0.55 x 106 x (120 - m) = 1000 m

=> 6996 - 58.3 m = 1000 m

=> m = 6.61 g

Therefore, we add 6.61 g of Na2CO3

c)

Mass of Solution = 1300 x 1.16 (Vlume (mL) x density)

=> Mass of Solution = 1508 g

=> Mass of Pb(NO3)2 = 0.15 x 1508 = 226.2 g

Therefore, we need 226.2 g of Pb(NO3)2

d)

Moles of Ba(OH)2 = 6.5 / 171.34 = 0.038 moles

Therefore,

Moles of HCl required = 0.038 x 2 = 0.076 moles

Molarity of HCl = 0.5 M

=> Volume of HCl required = 0.076 / 0.5 = 0.152 L = 152 mL

We need to prepare this HCl from 6 M HCl

Moles of HCl = 0.076

Moles = Molarity x Volume

=> 0.076 = 6 x V

=> V = 0.012667 L =12.67 mL

We will take 12.67 mL of 6 M HCl and add water to make final volume of 152 mL