Question #4fe76

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Answer 1 · 2017-05-16T22:22:50

Let the volume, $V = 1 L$ $V = 1"L"$ , then we know that the mass, $m = 1.2 g$ $m = 1.2"g"$ .

Find number of moles using the Ideal Gas Law:

$P V = n R T$ $PV = nRT$

where $P = 745 Torr$ $P = 745"Torr"$ , $T = 293^{\circ} K$ $T= 293^@"K"$ and $R = 62.364 \frac{LTorr}{Kmol}$ $R = 62.364 "L Torr"/"K mol"$

Solve the equation for n:

$n = \frac{P V}{R T}$ $n = (PV)/(RT)$

Substitute in the know values:

$n = \frac{(745 Torr) (1 L)}{(62.364 \frac{LTorr}{Kmol}) (293^{\circ} K)}$ $n = ((745"Torr")(1"L"))/((62.364 "L Torr"/"K mol")(293^@"K"))$

The molecular mass is

$\frac{m}{n}$ $m/n$

$29.4 \frac{g}{mol}$ $29.4"g"/"mol"$