Question

A beaker contains 20.0 ml of a 50.0 g/L solution. What will the new concentration be if you add 40.0 mL of water?

Answer 1

Answer:

moles SO42- = 0.0500 L x 0.20 M=0.010

moles Ba2+ = 0.0500 L x 0.10 M = 0.0050

Ba2+ + SO42- = BaSO4 (s)

moles SO42- in excess = 0.010 - 0.0050=0.0050

total volume = 100 mL = 0.100 L

[SO42-]= 0.0050/0.100= 0.050 M

[Na2SO4] = 2 /2 = 1 M

moles Na2SO4 = 2 M x 0.500 L = 0.500

mass Na2SO4 = 0.500 x 142 g/mol=71.0 g

moles MgBr2 = 46 /184 =0.25

moles Br- = 0.25 x 2 = 0.50

[Br-]= 0.50 / 0.50 L = 1 M