Answer:
The volume of 55 grams of BF₃ is 18.1664 L.
Explanation:
If the molar mass of compound BF₃ is 67.81 [tex]\frac{grams}{mole}[/tex], then the number of moles 55 grams of compound contains is calculated as:
[tex]55 grams* \frac{1 mole}{67.81 grams} = 0.811 moles[/tex]
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
Then it is possible to apply the following rule of three: if by definition of STP 1 mole of a compound occupies 22.4 L, 0.811 moles of the compound occupies how much volume?
[tex]volume=\frac{0.811 moles* 22.4 L}{1 mole}[/tex]
volume= 18.1664 L
The volume of 55 grams of BF₃ is 18.1664 L.
