Yes, this is a rounding error.

Where you calculate:

$$n = \frac{PV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = 0.18$$

The actual unrounded answer is $\pu{0.181072885 mol}$ - a little bit higher than the rounded value.

So, in your final equation:

$$0.181072885 \times 207.1 = \pu{37.5 g}$$

I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.

Yes, this is a rounding error.

Where you calculate: $$n = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$

The actual unrounded answer is $\pu{0.181072885 mol}$ - a little bit higher than the rounded value.

So, in your final equation:

$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$

I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.

Yes, this is a rounding error.

Where you calculate:

$$n = \frac{PV}{RT} = \frac{0.5 \ \text{atm} \cdot 20.0 \ \text{L}}{0.08206 \cdot 673 \text{K}} = 0.18$$

The actual unrounded answer is $0.181072885$ - a little bit higher than the rounded value.

So, in your final equation:

$0.181072885 \times 207.1 = 37.5g$

I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.

Yes, this is a rounding error.

Where you calculate:

$$n = \frac{PV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = 0.18$$

The actual unrounded answer is $\pu{0.181072885 mol}$ - a little bit higher than the rounded value.

So, in your final equation:

$$0.181072885 \times 207.1 = \pu{37.5 g}$$

I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.