Answer :
Final answer:
The volume of 15.6 g of H₂O(g) at 36.2°C and 1.25 atm is 2.06 L.
Explanation:
To find the volume of 15.6 g of H2O(g) at 36.2°C and 1.25 atm, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for volume V, we have V = (nRT) / P. First, we need to convert the mass of H2O(g) to moles. The molar mass of H2O is 18.015 g/mol, so the number of moles (n) is 15.6 g / 18.015 g/mol = 0.866 mol.
Next, we can substitute the values into the equation: V = (0.866 mol)(0.0821 L·atm/mol·K)(36.2°C + 273.15 K) / 1.25 atm = 2.06 L.
Therefore, the volume that 15.6 g of H2O(g) occupies at 36.2°C and 1.25 atm is 2.06 L.
