Determine the boiling point of a solution that contains 70.6 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.

Boiling point of the solution is 82.3°C

Explanation:

Boiling point elevation is defined as the increasing of a boiling point of a substance by the addition of a solute. The formula is:

ΔT = K×m×i

Where ΔT is change in temperature (Final T - 80.1°C), K is boiling point elevation constant (2.53°C/m), m is molality of the solution (moles of naphthalene / kg of benzene) and i is Van't Hoff factor (1 for Naphthalene)

Moles of 70.6g of naphthalene are:

70.6g × (1mol / 128.16g) = 0.5509 moles

Kg of 722mL of benzene are:

722mL × (0.877g / mL) × (1kg / 1000g) = 0.633kg of benzene

Replacing in boiling point elevation formula:

(T - 80.1°C) = 2.53°C/m×(0.5509mol / 0.633kg)×1

T - 80.1°C = 2.2°C

T = 80.1°C + 2.2°C

T = 82.3°C

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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