1. Describe how you would prepare each of the following aqueous solutions. a. 1.
ID: 982303 • Letter: 1
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.
Explanation / Answer
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
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