What is the molar volume of pure water at 22 degrees Celsius?
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To determine the molar volume of pure water at 22 degrees Celsius, we need to know the molar mass of water and use the ideal gas law equation.
First, let's find the molar mass of water (H2O). The molar mass of hydrogen (H) is approximately 1 g/mol, and the molar mass of oxygen (O) is approximately 16 g/mol. Since water consists of two hydrogen atoms and one oxygen atom, the molar mass of water is:
Molar mass of water (H2O) = (2 * molar mass of hydrogen) + molar mass of oxygen
= (2 * 1 g/mol) + 16 g/mol
= 2 g/mol + 16 g/mol
= 18 g/mol
The ideal gas law equation is:
PV = nRT
where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/(K.mol))
T = temperature (in Kelvin)
To convert the temperature from Celsius to Kelvin, use the formula:
T(K) = T(C) + 273.15
For 22 degrees Celsius:
T(K) = 22 + 273.15
= 295.15 K
Now that we know the molar mass of water and the temperature in Kelvin, we can rearrange the ideal gas law equation to solve for V:
V = (nRT) / P
Since we want to find the molar volume, V/n, we can divide both sides of the equation by n:
V/n = (RT) / P
Substituting the values into the equation:
V/n = (0.0821 L.atm/(K.mol)) * (295.15 K) / P
Assuming the pressure is at normal atmospheric pressure (1 atm), we can calculate the molar volume of pure water at 22 degrees Celsius as:
V/n = (0.0821 L.atm/(K.mol)) * (295.15 K) / 1 atm
Calculating the expression:
V/n = 24.43 L/mol
Therefore, the molar volume of pure water at 22 degrees Celsius is approximately 24.43 L/mol.