Question

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.

benzene boiling point=80.1 Kb=2.53

carbon tetrachloride boiling point=76.8 Kb=5.03

Answer 1

Answer:

Concentration = 1.32 m

Boiling point = $83.44^{0}C$

Explanation:

Let the boiling point of two solutions = T

Let the molal concentration of the two solutions= m

The relation between elevation in boiling point and molal concentration is:

$elevationinboiling point=K_{b}Xmolality$

i) For benzene

$T-80.1=2.53Xm$

$T=2.53m+80.1$ ....(1)

ii) For carbon tetrachloride

$T-76.8=5.03Xm$

$T=5.03Xm+76.8$ ...(2)

Equating the two equations:

$5.03Xm+76.8=2.53m+80.1\\m=1.32molal$

Putting value of "m" in equation "1"

$T=2.53(1.32)+80.1=83.44^{0}C$

Answer 2

Molal concentration, $m = 1.32 m$

Boiling point, $T_b =83.44^oC$

Boiling-point elevation:

It describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

The formula for boiling point elevation is:

$\triangle T_b=K_b*m$............(a)

i) For benzene

$\triangle T_b=K_b*m\\\\T-80.1=2.53 *m\\\\T= 80.1+2.53*m$............(1)

ii) For carbon tetrachloride

$\triangle T_b=K_b*m\\\\T-76.8=5.03 *m\\\\T= 76.8+5.03*m$............(2)

On equating equations 1 and 2:

$80.1+2.53*m=76.8+5.03*m\\\\m=1.32m$

Thus, value of molal concentration is 1.32m.

Now, substituting value for m in equation a:

$\triangle T_b=K_b*m\\\\T_b=2.53*1.32+80.1\\\\T_b=83.44^oC$

Find more information about Boiling point elevation here:

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