Respuesta :
Answer:
a) 70.074 of gallons of 0.05 M solution we can make.
b) 0.0834 M is the molar concentration of the resulting solution.
Explanation:
Concentration of alum solution = c
Moles of alum in solution = n
Volume of solution = V
[tex]c=\frac{n}{V(L)}[/tex]
a) Mass of alum = 10 lbs = 4535.92 g
(1 lb = 453.592 g )
Molar mass of alum[tex]Al_2(SO_4)_3 = 342 g/mol[/tex]
Moles of alum = [tex]\frac{4535.92 g}{342 g/mol}=13.263 mol[/tex]
Volume of alum solution of 0.05 m = V
[tex]0.05M=\frac{13.263 mol}{V}[/tex]
[tex]V=\frac{13.263 mol}{0.05 M}=265.26 L = 70.074 gallons[/tex]
(1 L = 0.264172 gallons)
70.074 of gallons of 0.05 M solution we can make.
b) Moles of alum = [tex]\frac{4535.92 g}{342 g/mol}=13.263 mol[/tex]
Volume of the barrel = 42 gallons = 158.99 L
Concentration of alum : C
[tex]C=\frac{13.263 mol}{158.99 L}=0.0834 mol/L=0.0834 M[/tex]
0.0834 M is the molar concentration of the resulting solution.
