Question

What is the density of one mole of He (molar mass \(=4\ g\ mol^{-1}\)) at \(300\ K\) and a pressure of \(0.82\ atm\)? \((R=0.082\ L\ atm\ mol^{-1}\ K^{-1})\)

• A. \(1.33\times10^{-2}\ g\ mL^{-1}\)
• B. \(1.33\times10^{-2}\ g\ L^{-1}\)
• C. \(1.33\times10^{-1}\ g\ L^{-1}\)
• D. \(1.33\times10^{-1}\ g\ mL^{-1}\)

Hint:

For gases, \[ d=\frac{PM}{RT} \] where \(d\) is density, \(P\) is pressure, \(M\) is molar mass, \(R\) is the gas constant and \(T\) is temperature.

Answer 1

The Correct Option is C

Step 1: Write density directly from the ideal gas law.
Since $PV=nRT$ and $n=\dfrac{mass}{M}$, combining both gives density in a single step:
\[ d=\frac{PM}{RT} \]

Step 2: Substitute the given values.
\[ d=\frac{0.82\times4}{0.082\times300} \]

Step 3: Simplify.
\[ d=\frac{3.28}{24.6}=0.1333\ g\,L^{-1} \]

Step 4: Write in scientific notation.
\[ \boxed{1.33\times10^{-1}\ g\,L^{-1}} \]
which matches option (3).