Question

Find the mass of oxygen gas with which $882\times10^{23}$ degrees of freedom are associated at N.T.P. Given: Molar mass of diatomic gas, oxygen is $32\text{ g mol}^{-1}$ and oxygen molecule possess three translational and two rotational degrees of freedom.

• A. $16\text{ g}$
• B. $2\text{ g}$
• C. $32\text{ g}$
• D. $5\text{ g}$

Hint:

Notice the arithmetic connection in the numbers: $12.044$ is exactly double Avogadro's constant factor $6.022$. Recognizing these numeric multiples early during long multiplication steps helps simplify fractions quickly and leads straight to $2\text{ g}$.

Answer 1

The Correct Option is B

Concept: The total number of degrees of freedom ($f_{\text{total}}$) present within a collection of molecules is equal to the number of degrees of freedom possessed by a single individual molecule ($f_{\text{molecule}}$) multiplied by the total number of molecules ($N$) in the sample: \[ f_{\text{total}} = N \cdot f_{\text{molecule}} \] The total number of molecules can be linked to the total mass using Avogadro's number ($N_A = 6.023 \times 10^{23}$) and the molar mass ($M$): $N = \left(\frac{m}{M}\right)N_A$.

Step 1: Calculate the total number of molecules ($N$) in the sample.
The problem states that a single diatomic oxygen molecule possesses $3$ translational and $2$ rotational modes, giving: \[ f_{\text{molecule}} = 3 + 2 = 5 \] Using the given total degrees of freedom $f_{\text{total}} = 1.882\times10^{23}$: \[ 1.882\times10^{23} = N \times 5 \implies N = \frac{1.882\times10^{23}}{5} = 0.3764\times10^{23}\text{ molecules} \]

Step 2: Convert the molecule count into mass units (grams).
Using the mole definition relationship: \[ N = \frac{m}{M} \cdot N_A \implies m = \frac{N \cdot M}{N_A} \] Substitute $N = 0.3764\times10^{23}$, $M = 32\text{ g mol}^{-1}$, and $N_A = 6.023 \times 10^{23}$: \[ m = \frac{(0.3764\times10^{23}) \times 32}{6.023 \times 10^{23}} = \frac{0.3764 \times 32}{6.023} \] \[ m = \frac{12.0448}{6.023} \approx 1.9998\text{ g} \approx 2\text{ g} \]