The concentration of the mixture obtained by mixing two solutions, a and b, is 20%. the concentration of the second mixture obtained by mixing the same two solutions is 14.8%. find the concentration of each of the solutions a and b, if the first mixture is obtained by mixing them in 1: 1 ratio and the second is obtained by mixing them in 3: 7 ratio.

Concentration of solution A is 33% and B is 7%.

Step-by-step explanation:

Let the concentration of solution A be C1 and concentration solution B is C2.

Now we start the question.

If these two solutions are mixed in the ratio of V1 and V2.

Now the molecules present in the solution by mixing these two solutions will be= Total molecules of solution A + solution B

So the first equation will be C'V' =C1V1+C2V2

Here C'=20% and V1:V2=1:1 and V'=V1+V2

.20×(V+V) = C1×V+C2V

⇒0.20×2V= V(C1+C2)

⇒0.40=C1+C2(1)

Now other solution is prepared in which

C"V"=C1V3+C2V4

Here C"=14.8% V"=(V3+V4) & V3:V4=3:7

⇒.148×10V = C1×3V+C2×7V

⇒1.48V= V(3C1+7C2)

⇒1.48 = 3C1+7C2 (2)

Now we multiply equation (1) by 3 then

3C1+3C2= 0.40×3=1.20(3)

Now we subtract equation (3) from (2)

7C2-3C2=1.48-1.20

4C2 =0.28

C2 = .07 or we can say concentration of solution B is 7%

Now we put the value of C2 in equation (1)

C1+.07 = 0.40

C1 = 0.40-0.07 = 0.33 or concentration of solution A is 33%.

x= 149. 24+7=31. 180-31=149

your welcome

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation: