[administration]

Q. There are two solutions of sulphuric acid (acid+water) with concentrations of 50% and 80% respectively. They are mixed in a certain ratio to get a 62% sulphuric acid sol. This sol. is mixed with 6lit of water to get back 50% sol. how much 80% sol. has been used in the entire process?

[Wonder Boy]

[sol1] = 50%

[sol2] = 80%

now formula for mixture problem is new conc c3 = [c1 . n1 + c2. n2]/[n1+n2]

where c1 & c2 are concentrations of the 2 solutions and n1 & n2 are the amounts(in lit.) of the solutions with conc 50% & 80% respectively. They are mixed to give sol. of conc. c3

 

so 0.62 = [.50xn1 + .80 x n2] /[n1 + n2]

–> 0.62 n1 + 0.62 n2 = .5 n1 + .8 n2

–> .12 n1 = .18 n2 –> 2 n1 = 3 n2 –> n1/n2 = 3/2 ………(i)

Also let n3=(n1 + n2) litres of 62 % acid sol. is mixed with 6 lit of water (pure water is used so concentration of it is 0% as it doesnt contain any sulphuric acid).

 

using the forumla again  .50 = [.62 x n3 + 0 x 6] /[n3 + 6]

.5 n3 + 3 = .62 n3

–> .12 n3 = 3 –> n3 = 300/12 –> n3 = 25 lit.

–> n1 + n2 = 25

 

putting n1= (3/2).n2  from (i)

(3/2).n2  + n2 = 25

–> 5/2 . n2 = 25 –> n2 = 10 lit.

So 10 lit. of 80% solution has been used in  entire process.

[Author]

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