A200 ml sample of 0.1015 miric acid is mbred with 2300 ml of water. what is the molar concentration of nitric acid in the final solution 0 406 m 127m 325 x 10 m 5.08 x 10 m 8.12 x 10 m water backspace lu 'o pin

Molar concentration of nitric acid in the final solution is

Explanation:

According to the dilution law,

where,

= molarity of stock solution = 0.1015 M

= volume of stock solution = 200 ml

= molarity of dilute solution = ?

= volume of  dilute  solution = (2300 +200 )ml = 2500 ml

Putting in the values we get:  

Thus the molar concentration of nitric acid in the final solution is

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