[tex]\tt{\huge{\red{Solution:}}}[/tex]
Step 1: List the given values.
[tex]\begin{aligned} P & = 779 \: \cancel{\text{mmHg}} \times \frac{\text{1 atm}}{760 \: \cancel{\text{mmHg}}} = \text{1.025 atm} \\ T & = 40^{\circ}\text{C} = \text{313.15 K} \\ \text{MM} & = \text{86.172 g/mol} \end{aligned}[/tex]
Final Step: Calculate the density of a certain hydrocarbon.
[tex]\begin{aligned} d & = \frac{(P)(\text{MM})}{RT} \\ & = \frac{(\text{1.025 atm})(\text{86.172 g/mol})}{\left(0.082057 \: \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(\text{313.15 K})} \\ & = \boxed{\text{3.44 g/L}} \end{aligned}[/tex]
Hence, the density of a certain hydrocarbon at 40°C and 779 mmHg is 3.44 g/L.
[tex]\\[/tex]
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