Question

You have 95.0 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 2.50 M solution of AgNO3(aq). Calculate the concentration of Na+ after the two solutions are mixed together.
1.08 M
0.475 M
5.00 M
0.00 M
2.16 M

Answer 1

mols Na from Na2CrO4 = 0.095 x 2.50 = ?
mols Na from AgNO3 = 0
total volume = (95 mL + 125 mL)/1000) = ?L
(Na) = mols Na/total volume in L.

Answer 2

mols Na from Na2CrO4 = 0.095 x 2.50 = .2375

mols Na from AgNO3 = 0

total volume = (95 mL + 125 mL)/1000) = .22L

(Na) = mols Na/total volume in L. .2375/.22L= 8.1.07954M

Answer 3

mols Na from Na2CrO4 = 0.095 x 2.50 = .2375

mols Na from AgNO3 = 0

total volume = (95 mL + 125 mL)/1000) = .22L

(Na) = mols Na/total volume in L. .2375/.22L= 1.07954M

Answer 4

Dr.Bob, 1.07954M or 1.08M isn't the answer but 2.16M is. so, why is that. You make a mistake there Dr.Bob222. Can you solve the problem again?

Answer 5

1st: 95mL÷1000mL=.090L because there are 1000mL in liter
2nd: .095mL*2.50M=.2375M per L
3rd: .2375M per Liter *2 =.475M per L because we have two Na(Na2)
4th: 95mL+125mL= 220mL
5th: 220mL÷1000=.220L
6th: .475M per L÷.220L=2.159M
Answe: Your answeris e which is 2.16M