Question #13f28

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Answer 1 · 2017-05-13T04:50:12

$n = \frac{volume}{molar volume} = \frac{0.442}{22.7} = 0.0195 mol$ $n="volume"/"molar volume"=0.442/22.7=0.0195" mol"$

$molar mass = \frac{mass}{n} = \frac{2.87}{0.0195} = 147 g {mol}^{- 1}$ $"molar mass"="mass"/"n"=2.87/0.0195=147" g"" mol"^-1$

STP = standard temperature and pressure

P = 100 kPa
T = 273.15 K

Assuming an ideal gas, it will obey the Ideal Gas Law:

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

Rearranging, we get

$V = \frac{nRT}{P}$ $V="nRT"/"P"$

Where R is the Gas Constant

$R = 8.314 J K^{- 1} {mol}^{- 1}$ $R=8.314" J " "K"^-1 " mol"^-1$

Using the above value for the gas constant (R) and pressure in kPa, the volume will come out with units of L.

We can use the Ideal Gas Law to solve for the volume that 1 mole of any ideal gas occupies at 273.15 K and 100 kPa:

$molar volume = \frac{1 \cdot 8.314 \cdot 273.15}{100} = 22.7 L$ $"molar volume"=(1*8.314*273.15)/100=22.7" L"$

Next, we find the number of moles of gas that we have by dividing the known volume by the molar volume.

$n = \frac{volume}{molar volume} = \frac{0.442}{22.7} = 0.0195 mol$ $n="volume"/"molar volume"=0.442/22.7=0.0195" mol"$

Finally, rearrange the mole formula to solve for the molar mass.

$n = \frac{mass}{molar mass}$ $n="mass"/"molar mass"$

$molar mass = \frac{mass}{n} = \frac{2.87}{0.0195} = 147 g {mol}^{- 1}$ $"molar mass"="mass"/"n"=2.87/0.0195=147" g"" mol"^-1$