Verified answer

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & P = \text{733 mmHg = 0.96447 atm} \\ & T = 46.0^{\circ}\text{C} = \text{319.15 K} \\ & MM = \text{80.912 g/mol} \end{aligned}[/tex]

Step 2: Calculate the density.

[tex]\begin{aligned} d & = \frac{(P)(MM)}{RT} \\ & = \frac{(\text{0.96447 atm})(\text{80.912 g/mol})}{\left(0.082057 \: \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(\text{319.15 K})} \\ & = \boxed{\text{2.98 g/L}} \end{aligned}[/tex]

Hence, the density of hydrogen bromide at 46.0°C is 2.98 g/L.

[tex]\\[/tex]

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