# A sample of gas weighing 0.0286 g occupies a volume of 50 cm^3 at 76 cmHg. and 25 degree centigrade. What is the molar mass of the gas?

Jun 26, 2017

$M M = 14.0$ $\text{g/mol}$

#### Explanation:

We can find the molar mass $M M$ of this gas using the equation

$M M = \frac{\mathrm{dR} T}{P}$

where

• $d$ is the density of the gas, in $\text{g/L}$

• $R$ is the universal gas constant, equal to $0.082057 \left(\text{L"•"atm")/("mol"•"K}\right)$

• $T$ is the absolute temperature of the gas, in $\text{K}$

• $P$ is the pressure of the gas, in $\text{atm}$

We need to convert some units to get where we need to be. Let's do the density first.

$1$ $\text{L}$ is equal to $1$ ${\text{dm}}^{3}$, which is equal to ${10}^{3}$ ${\text{cm}}^{3}$, so the density is

((0.0286color(white)(l)"g")/(50cancel("cm"^3)))((10^3cancel("cm"^3))/(1cancel("dm"^3)))((1cancel("dm"^3))/(1color(white)(l)"L")) =color(red)(0.572 color(red)("g/L"

Degrees centigrade is the same as degrees Celsius, so let's convert this to $\text{K}$ by adding $273$:

T = 25^"o""C" + 273 = color(green)(298 color(green)("K"

Lastly, let's convert our pressure, which is in centimeters of mercury:

76cancel("cm Hg")((10cancel("mm Hg"))/(1cancel("cm Hg")))((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(purple)(1 color(purple)("atm"

Now that we have all our values, let's plug them into the equation:

MM = ((color(red)(0.572"g"/"L"))(0.082057("L"•"atm")/("mol"•"K"))(color(green)(298"K")))/(color(purple)(1"atm"))

 = color(blue)(14.0 color(blue)("g/mol"