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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.

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    $\begingroup$ You have mass fraction, that is what you need. Mass alone is not useful. Molar mass of unknown then relates mass fraction and molality. $\endgroup$ – Poutnik Dec 7 '19 at 13:12
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    $\begingroup$ 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. $\endgroup$ – Maurice Dec 7 '19 at 21:44
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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.

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