Question

Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is k % acid. From jar C, 2/3 liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Find k.

Answer 1

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added  to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.