1. Use the formula for translational kinetic energy of an ideal gas

For all molecules of an ideal gas, the total kinetic energy of translation is

$K = \frac{3}{2} nRT$

where:

• $n$ = number of moles
• $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$
• $T$ = absolute temperature

2. Calculate number of moles of $\text{CO}_2$

Given mass of $\text{CO}_2$:

$m = 50\ \text{g}$

Molar mass of $\text{CO}_2$:

$M = 44\ \text{g mol}^{-1}$

So,

$n = \frac{m}{M} = \frac{50}{44} = 1.13636\ \text{mol}$

3. Convert temperature into kelvin

$T = 17^{\circ}\text{C} + 273 = 290\ \text{K}$

(Using $290\,\text{K}$ is standard here.)

4. Substitute into the formula

$K = \frac{3}{2} \times \frac{50}{44} \times 8.314 \times 290$

First compute:

$8.314 \times 290 = 2411.06$

Then,

$\frac{50}{44} \times 2411.06 = 2739.84$

Now,

$K = \frac{3}{2} \times 2739.84 = 4109.76\ \text{J}$

Using the common approximation $R = 8.31$:

$K = \frac{3}{2} \times \frac{50}{44} \times 8.31 \times 290$

$= 4108.28\ \text{J}$

This is closest to Option D.

5. Check options

• A: $4205.5\,\text{J}$ — not matching
• B: $3582.7\,\text{J}$ — too low
• C: $3986.3\,\text{J}$ — not matching
• D: $4102.8\,\text{J}$ — closest and correct by standard rounding/approximation

6. Final answer

The kinetic energy of translation of the molecules is

$\boxed{4102.8\ \text{J} \text{ (approximately)}}$

So the correct option is D.