Recall that **molecular weight** is in grams per 1 mole of a substance.

$\overline{){\mathbf{M}}{\mathbf{o}}{\mathbf{l}}{\mathbf{e}}{\mathbf{c}}{\mathbf{u}}{\mathbf{l}}{\mathbf{a}}{\mathbf{r}}{\mathbf{}}{\mathbf{W}}{\mathbf{e}}{\mathbf{i}}{\mathbf{g}}{\mathbf{h}}{\mathbf{t}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{W}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{g}}{\mathbf{m}\mathbf{o}\mathbf{l}}}$

First, we have to calculate the amount of gas in **moles** using the **ideal gas equation**.

$\overline{){\mathbf{P}}{\mathbf{V}}{\mathbf{=}}{\mathbf{n}}{\mathbf{R}}{\mathbf{T}}}$

P = pressure, atm

V = volume, L

n = moles, mol

R = gas constant = 0.08206 (L·atm)/(mol·K)

T = temperature, K

**Isolate n (number of moles of gas): **

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below 100 ^{o}C in a boiling-water bath and determine the mass of vapor required to fill the bulb (see drawing

).

From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, 1.013 g ; volume of bulb, 354 cm^{3} ; pressure, 743 torr ; temperature, 99 ^{o}C .

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