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: Use the ideal gas equation.
For one mole of an ideal gas, \[ PV=nRT \] Given, \[ n=1,\quad P=0.82\ atm,\quad R=0.082\ L\,atm\,mol^{-1}K^{-1}, \quad T=300\ K \]

Step 2: Calculate the volume occupied by one mole of helium.
Using \[ V=\frac{nRT}{P} \] \[ V=\frac{1\times0.082\times300}{0.82} \] \[ V=\frac{24.6}{0.82} \] \[ V=30\ L \]

Step 3: Calculate density.
Density is given by \[ d=\frac{\text{Mass}}{\text{Volume}} \] For one mole of helium, \[ \text{Mass}=4\ g \] Therefore, \[ d=\frac{4}{30} \] \[ d=0.133\ gL^{-1} \]

Step 4: Express in scientific notation.
\[ 0.133 = 1.33\times10^{-1} \] Thus, \[ d=1.33\times10^{-1}\ gL^{-1} \]

Step 5: Final conclusion.
Hence, \[ \boxed{1.33\times10^{-1}\ gL^{-1}} \] which corresponds to option (3).