How many quarts of pure antifreeze must be added to 8 quarts of a 40% antifreeze solution to obtain a 60% antifreeze solution?

we have 8 quarts of 40% antifreeze,

according to the britain equivalent, 1 quart = 1.13L

so, we have 8 * 1.13L  = 9.04 L

out of the 9.04 L, 40% is antifreeze

amount of antifreeze = 0.4 * 9.04 = 3.616 L

moles of antifreeze present = 3.616/22.4 = 0.16 moles

moles of antifreeze in 60% solution = 0.6 * v/22.4 = 0.027 * v          

(where v is the volume of final solution)

molarity of initial solution = 0.16/9.04 = 0.018 M

molarity of final solution = 0.027*v/v = 0.027 M

m1*v1  = m2*v2

0.018 * 9.04 = 0.027 * v

v = 0.018 * 9.04/0.027

v = 6.026 L

amount if antifreeze in the final solution = 6.026 * 0.027 =  

0.163 Moles of antifreeze

amount in L = 0.163 * 22.4 = 3.65 L

amount in quarts  = 3.65/1.13 = 3.23 quarts