What is the freezing point of a solution prepared by dissolving 50.0g of glycerin in 200g of ethanol?

-120 °C

Explanation:

The freezing point depression is calculated as follows:

ΔT = Kf*m

where ΔT is the difference between the freezing point of the pure solvent and from the solution; kf is a constant (equal to 2 °C/m for ethanol); and m is the molality of the solution.

moles of solute = (50 g)/(92.09 g/mol) = 0.54 mol

molality of the solution = moles of solute / kg of solvent

molality of the solution = 0.54 mol/ 0.2 kg = 2.7 m

ΔT = 2*2.7 = 5.4 °C

The freezing point of ethanol is -114.6 °C, then the freezing point of the solution is -114.6 °C - 5.4 °C =  -120 °C

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