A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving some potassium bromide () in of . This solution boils at . Calculate the mass of that was dissolved.

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

Best regards.

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