91Ó°ÊÓ

A substance's volume is the amount of space it occupies.

Let us solve the given problem.

We have,

\(PV = nRT\)

where,

\(P = \)pressure of the gas.

\(V = \)volume of the gas.

\(n = \)number of moles of the gas.

\(T = \)temperature of the gas.

\(\begin{aligned}{}R &= {\rm{ Ideal gas constant}}\\ &= 0.08206\;{\rm{L atm mo}}{{\rm{l}}^{ - 1}}{K^{ - 1}}\end{aligned}\)

If, amount of the gas\((n)\)is in moles, temperature\((T)\)is in Kelvin\((K)\), pressure\((P)\)is in atm.

\(\begin{aligned}{}\frac{P}{{RT}} &= \frac{n}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{{n \cdot (\mu )}}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{m}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \rho \end{aligned}\)

Density\((\rho ) = \frac{{P \cdot (\mu )}}{{RT}}\)

Where\(\mu = \)molar mass.

Consider the given problem.

Mass\((m) = 0.0494\) grams.

Volume \((V) = 0.1\;{\rm{L}}\).

Pressure \((P) = 307\) torr.

Temperature \((T) = {26^\circ }.C\)

Converting the temperature in degree Celsius to kelvins.

We have, \({0^\circ }C = (0) + 273.15K\)

So,

\(\begin{aligned}{}{28^\circ }C = (28) + 273.15K\\ = 301.15K\end{aligned}\)

Converting pressure in torr into atm.

We have, 1 torr \( = 0.0013\;{\rm{atm}}\).

So,

\(\begin{aligned}{}307{\rm{ torr}} &= (307) \cdot (0.0013){\rm{ atm}}{\rm{. }}\\ &= 0.4\;{\rm{atm}}.\end{aligned}\)

Calculate total molar mass.

\(\begin{aligned}{}\frac{{P \cdot (\mu )}}{{RT}}& = \frac{m}{V}{\rm{ }}\\{\rm{molar mass }}(\mu )& = \frac{{mRT}}{{PV}}\\ &= \frac{{(0.0494) \cdot (0.08206) \cdot (301.15)}}{{(0.4) \cdot (0.1)}}\\ &= \frac{{1.22}}{{0.04}}\\ &= 30.5\frac{{{\rm{gram}}}}{{{\rm{mol}}}}\end{aligned}\)

Therefore, calculated molar mass is \((\mu ) = 30.5\frac{{{\rm{gram}}}}{{{\rm{mol}}}}\).