The boiling point of benzene C 6 H 6 is 80.10 o C at 1 atmosphere. A nonvolatile

Question

The boiling point of benzene C6H6 is 80.10oC at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is estrogen (estradiol).

If 10.77 grams of estrogen, C18H24O2 (272.4 g/mol), are dissolved in 244.3 grams of benzene ...

The molality of the solution is m.

The boiling point of the solution is 0C.

The freezing point of ethanol CH3CH2OH is -117.300C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in ethanol is chlorophyll .

How many grams of chlorophyll, C55H72MgN4O5 (893.5 g/mol), must be dissolved in 231.0 grams of ethanol to reduce the freezing point by 0.4000C ?

g chlorophyll.

Explanation / Answer

We need to use the formula

change in boiling point = Kb* molality

please provide the Kb values for both the problems.

From the book , I have taken the Kb & Kf values.

Kb for benzene = 2.53 C Kg / mol

Kf for ethanol = 1.99 C Kg / mol

a) Benzene + estrogen

no.of moles of estrogen = 10.77/ 272.4 = 0.0395374449339207

molality = no.of moles of estrogen / weight of solvent ( in kg ) = 0.03953/0.2443 = 0.1618 m

molality= 0.1618 m.

change in boiling point = Kb* molality

change in boiling point = 2.53 *0.1618 = 0.409354 C

new boiling point = 80.1 + 0.409354 = 80.509354 C.

b) ethanol + chlorophyll

Here change in freezing point = 0.4

so, change in freezing point = Kf* molality

0.4 = 1.99 * molality

molality = 0.201

molality = no.of moles of chlorophyll / weight of solvent ( in kg )

0.201 = no.of moles of chlorophyll /0.231

no.of moles of chlorophyll = 0.04643

weight of chlorophyll = no.of moles * molecular weight = 0.04643 * 893.5 = 41.485205 gms.

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