Question

A gaseous product of a reaction is collected at 291K and 0.96 atm. Given R=0.0821L⋅atmmol⋅K
, what is the molar mass of the gas, in grams per mole, when 3.12 g of gas occupy 4.85 L?(1 point)
Responses

17.9
17.9

38.7
38.7

15.4
15.4

16.0

Answer 1

To find the molar mass of the gas, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) is the pressure (in atm),
- \( V \) is the volume (in L),
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (\( 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)),
- \( T \) is the temperature (in K).

First, we will solve for \( n \):

Rearranging the equation gives:

\[ n = \frac{PV}{RT} \]

Now we can substitute the known values into the equation:

- \( P = 0.96 \, \text{atm} \)
- \( V = 4.85 \, \text{L} \)
- \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)
- \( T = 291 \, K \)

Substituting these values in:

\[
n = \frac{(0.96 \, \text{atm})(4.85 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(291 \, K)}
\]

Calculating the right side:

1. Calculate the numerator:
\[
(0.96)(4.85) = 4.656 \, \text{atm} \cdot \text{L}
\]

2. Calculate the denominator:
\[
(0.0821)(291) = 23.92 \, \text{L} \cdot \text{atm} / \text{mol}
\]

Now, calculate \( n \):

\[
n = \frac{4.656}{23.92} \approx 0.194 \, \text{mol}
\]

Now we can find the molar mass \( M \) using the formula:

\[
M = \frac{\text{mass}}{n}
\]

Where the mass of the gas is given as \( 3.12 \, g \):

\[
M = \frac{3.12 \, g}{0.194 \, \text{mol}} \approx 16.1 \, g/mol
\]

Among the options given, the closest value is:

**16.0 g/mol**.

So, the answer is **16.0**.