# College Chemistry - Raised Boiling Point [closed]

Determine the molar mass of an unknown if you dissolve enough of the unknown into Benzene to make a 1.55% mass percent mixture of unknown to benzene and find that the boiling point of Benzene was raised by 2.3%

For this question, I just have a problem understanding how to get the molar mass without having grams or moles given. Now I don't know if I have to chose my own grams or if there is another way to find the molar mass of the unknown.

• You have mass fraction, that is what you need. Mass alone is not useful. Molar mass of unknown then relates mass fraction and molality. – Poutnik Dec 7 '19 at 13:12
• The question is not clearly defined. Saying that the temperature is increased by 2.3% has no meaning. Which temperature scale ? The absolute temperature in Kelvin ? The usual temperature in centigrade degrees ? in Fahrenheit ? 2.3% of 0°C is zero. 2,3% of 273 K is about 5 K. That is not the same. – Maurice Dec 7 '19 at 21:44

This question has a lot of flaws to be a college chemistry question. Therefore, I decided to give some insight even though this is a clearly a homework question.

The question did not have enough data such as the ebullioscopic constant ($$K_b$$) of benzene and the van't Hoff factor ($$i$$) of the solute. At least you can find the ebullioscopic constant of benzene as $$\pu{2.53 K\:kg\:mol-1}$$ from Wikipedia. If we assume the van't Hoff factor ($$i$$) of the solute is unity, we can solve this problem without much difficulty. However, we also have to assume that the percent temperature elevation (although it is unitless in %) is given in Kelvin scale ($$\pu{K}$$) because of units in $$K_b$$. If it was given in Celsius ($$\pu{^\circ C}$$), this percentage would change.

Suppose molar mass of unknown is $$M_u$$. If the solution is 1.55% mass percent ($$w/w$$) mixture of unknown to benzene, the masses used are $$\pu{1.55 g}$$ of unknown and $$\pu{(100-1.55) g}=\pu{98.45 g}$$ of benzene to make $$\pu{100 g}$$ of solution. Therefore, molality of the solution ($$m_b$$) is: $$m_b = \frac{(1.55/M_u)\ \pu{mol}}{\pu{98.45 g}\times \frac{\pu{1 kg}}{\pu{1000 g}}} = \frac{1550}{98.45M_u} \pu{mol/kg}$$

Now we can apply the boiling point elevation: $$\Delta T = iK_b m_b$$, where $$K_b = \pu{2.53 K\:kg\:mol-1}$$ and we assumed $$i=1$$. Numerically:

$$2.53 \times \frac{1550}{98.45M_u} = \Delta T = \left(\frac{\Delta T}{\pu{353.2 K}} \times 100 \right) \times \frac{\pu{353.2 K}}{100} = 2.3 \times \frac{\pu{353.2 K}}{100}$$

$$\therefore \; M_u= 2.53 \times \frac{1550}{98.45} \times \frac{100}{2.3 \times 353.2 } = \pu{4.90 g/mol}$$

This is a awfully small number for a non-gaseous compound. The only wrong assumption should be the boiling point of benzene in Kelvin so let's make the switch to Celsius scale, which is $$\pu{80.1 ^\circ C}$$:

$$2.53 \times \frac{1550}{98.45M_u} = \Delta T = \left(\frac{\Delta T}{\pu{80.1 ^\circ C}} \times 100 \right) \times \frac{\pu{80.1 ^\circ C}}{100} = 2.3 \times \frac{\pu{80.1 ^\circ C}}{100}$$

$$\therefore \; M_u= 2.53 \times \frac{1550}{98.45} \times \frac{100}{2.3 \times 80.1 } = \pu{21.62 g/mol}$$

Which is a better number.