Asked by anthoony
calculate the amount of water that must be added to 6.5 grams of urea to produce a 22-2% by mass solution?
Answers
Answered by
DrBob222
6.50/g soln) = 0.222
g soln = 6.50/0.222 = 29.3 g soln
29.3g soln-6.5g urea = 22.8 g H2O .
check:
[6.50g urea/(6.5g urea + 22.8)]*100 =
[6.50g urea/(29.3g soln)]*100 = 22.18 which rounds to 22.2%
g soln = 6.50/0.222 = 29.3 g soln
29.3g soln-6.5g urea = 22.8 g H2O .
check:
[6.50g urea/(6.5g urea + 22.8)]*100 =
[6.50g urea/(29.3g soln)]*100 = 22.18 which rounds to 22.2%