Question

You have an-empty 42-gatbarrel-and-a 10-lb bag of dry alum, Al2(SO4). (a) Calculate the number of gallons of 0.05 M solution you can make with this amount of alumthe-materials you hve on hand. (b) Suppose you put all the alum into a 42-gal-the barrel and filled it full Calculate the molar concentration of the resulting solution

Answer 1

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.

Step-by-step explanation:

Concentration of alum solution = c

Moles of alum in solution = n

Volume of solution = V

`c=(n)/(V(L))`

a) Mass of alum = 10 lbs = 4535.92 g

(1 lb = 453.592 g )

Molar mass of alum
`Al_2(SO_4)_3 = 342 g/mol`

Moles of alum =
`(4535.92 g)/(342 g/mol)=13.263 mol`

Volume of alum solution of 0.05 m = V

`0.05M=(13.263 mol)/(V)`

`V=(13.263 mol)/(0.05 M)=265.26 L = 70.074 gallons`

(1 L = 0.264172 gallons)

70.074 of gallons of 0.05 M solution we can make.

b) Moles of alum =
`(4535.92 g)/(342 g/mol)=13.263 mol`

Volume of the barrel = 42 gallons = 158.99 L

Concentration of alum : C

`C=(13.263 mol)/(158.99 L)=0.0834 mol/L=0.0834 M`

0.0834 M is the molar concentration of the resulting solution.