Question

What is the molar mass of a gas whose density is 3.55 g ⋅ l 1 at 110 c and a pressure of 736 torr?

Answer 1

115.25g/mol

Explanations:

According to the ideal gas equation:

`PV=nRT`

where

• P is the ,pressure (,in atm)

,

• V is the ,volume ,in litres

,

• n is the ,number of moles

,

• R is the ,Gas constant

,

• T is the ,temperatur,e in kelvin

Recall that:

`\begin{gathered} moles=\frac{mass}{molar\text{ mass}} \\ n=(m)/(M) \end{gathered}`

Substitute into the ideal equation to have:

`\begin{gathered} PV=(m)/(M)\cdot RT \\ PM=(m)/(V)\cdot RT \\ Since\text{ }\rho=(m)/(V),hence; \\ PM=\rho RT \end{gathered}`

The expression for calculating the molar mass will be:

`M=(\rho RT)/(P)`

Given the following parameters

`\begin{gathered} density\text{ }\rho=3.55gL^(-1) \\ R=(0.08205Latm)/(molK) \\ T=110^0C+273=383K \\ P=736torr=0.968atm \end{gathered}`

Substitute the given parameters into the formula

`\begin{gathered} M=(3.55*0.08205*383)/(0.968) \\ M=(111.559)/(0.968) \\ M=115.25gmol^(-1) \end{gathered}`

Hence the molar mass of the gas will be approximately 115.25g/mol