0.4g of a gas has a volume of 227cm^3 at 27°C and at a pressure of 100kPa. Calculate the relative molecular mass of the gas (R=8.31 J/kmol)?

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0.4g of a gas has a volume of 227cm^3 at 27°C and at a pressure of 100kPa. Calculate the relative molecular mass of the gas (R=8.31 J/kmol)?

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Answer 1 · 2015-09-10T17:03:03

$M_{r} = 43.95$ $M_r=43.95$

Explanation:

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

$n = \frac{p V}{R T}$ $n=(pV)/(RT)$

$= \frac{100 \times 10^{3} \times 227 \times 10^{- 6}}{8.31 \times 300}$ $=(100xx10^3xx227xx10^(-6))/(8.31xx300)$

$= 0.0091$ $=0.0091$

$n = \frac{m}{M_{r}}$ $n=m/M_r$

$M_{r} = \frac{m}{n} = \frac{0.4}{0.0091} = 43.95$ $M_r=m/n=0.4/0.0091=43.95$

It looks like it could be $C O_{2}$ $CO_2$ but there are other possibilities.

Answer 2 · 2015-09-10T17:11:38

Molecular mass: $44 g/mol$ $"44 g/mol"$ .

Explanation:

To solve this problem you need to use the ideal gas law equation

$P V = n R T$ $color(blue)(PV = nRT)" "$ , where

$P$ $P$ - the pressure of the gas;
$V$ $V$ - the volume of th gas;
$n$ $n$ - the number of moles of gas;
$R$ $R$ - the universal gas constant, in your case given as $8.31 J/mol K$ $"8.31 J/mol K"$
$T$ $T$ - the temperature of the gas - expressed in Kelvin!

The trick here is to realize that the units of $R$ $R$ can be rewritten as

$R = \frac{J}{mol \cdot K} = \frac{kg \cdot m^{2}}{s^{2}} \cdot \frac{1}{mol \cdot K}$ $R = "J"/("mol" * "K") = ("kg" * "m"^2)/"s"^2 * 1/("mol" * "K")$

$= \frac{kg}{m \cdot s^{2}} \cdot m^{3} \cdot \frac{1}{mol \cdot K}$ $="kg"/("m" * "s"^2) * "m"^3 * 1/("mol" * "K")$

$= \frac{Pa \cdot m^{3}}{mol \cdot K}$ $= ("Pa" * "m"^3)/("mol" * "K")$

To find the molecular mass of the gas, you need to know how many moles of gas have a mass of 0.4 g. Rearrange the ideal gas law equation to solve for $n$ $n$

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

Next, convert the pressure from kPa to Pa and the volume from cubic centimeters to cubic meters

$100color(red)(cancel(color(black)("kPa"))) * "1000 Pa"/(1color(red)(cancel(color(black)("kPa")))) = 10^5"Pa"$

and

$227color(red)(cancel(color(black)("cm"^3))) * ("1 m"""^3)/(10^6color(red)(cancel(color(black)("cm"^3)))) = 227 * 10^(-6)"m"^3$

Plug in these values to find $n$ , but do not forget to convert the temperature from degrees Celsiu to Kelvin!

$n = (10^5color(red)(cancel(color(black)("Pa"))) * 227 * 10^(-6)color(red)(cancel(color(black)("m"^3))))/(8.31(color(red)(cancel(color(black)("Pa"))) * color(red)(cancel(color(black)("m"^3))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 27)color(red)(cancel(color(black)("K")))) = "0.00910 moles"$

The molar mass of the gas will be

$M_M = m/n = "0.4 g"/"0.00910 moles" = color(green)("44 g/mol")$

SIDE NOTE I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the gas.

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0.4g of a gas has a volume of 227cm^3 at 27°C and at a pressure of 100kPa. Calculate the relative molecular mass of the gas (R=8.31 J/kmol)?

2 Answers

Explanation:

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