There are 3 beakers each of which contains saline solution.
beaker a initially contains 3 liters of 10\% salt solution.
beaker b initially contains 2 liters of 20\% salt solution.
beaker c initially contains 4 liters of 0\% salt solution.

two liters are transferred from a to b and the result is thoroughly mixed. then one liter is transferred from b to c and the result mixed. finally two liters are transferred from c back to a. what is the percentage concetration of salt in a after all this?

percentage concentration of salt in a= \%

Explanation:

Assume all the concentrations are expressed in volumetric terms, i.e 10% = 10 liter salt / 100 liter solution, 20% = 20 liter salt / 100 liter soluton, 0% = 0 liters salt.

1) First transformation: 2 liters are from A to B

Solution A:

Solution B:

Resultant mixture in beaker B: 2 liters of solution A plus 2 liters of solution B

2) Second transformation: 1 liter transferred from B to C

                       ↑              ↑

                  (from B)     ( in C)

3) Third transformation: 2 liters are from C to A.

                                       ↑                                    ↑

                                (from C)                             (in A)

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml