In the determination of the molar mass of a volatile liguid a 250ml flask was found to contain 0.250g of the vapor measured at 273c and 760 mmhg pressure. The molar mass is nearest the value
Use PV = nRT and solve for n = number of mols. Then n = grams/molar mass. You know grams and moles, solve for molar mass.
To determine the molar mass of a volatile liquid, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles (in mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given measurements to the appropriate units. The pressure is already in mmHg, so we need to convert it to atm by dividing by 760 mmHg/atm.
P = 760 mmHg / 760 mmHg/atm
P = 1 atm
Next, the volume is given as 250 mL, so we need to convert it to liters by dividing by 1000 mL/L.
V = 250 mL / 1000 mL/L
V = 0.25 L
The temperature is given as 273°C, but we need to convert it to Kelvin by adding 273.15.
T = 273°C + 273.15
T = 546.15 K
Now, we can substitute these values into the ideal gas law equation:
(1 atm)(0.25 L) = n(0.0821 L·atm/(mol·K))(546.15 K)
Simplifying the equation:
0.25 = n(0.0821)(546.15)
0.25 = 44.860215n
Solving for n:
n = 0.25 / 44.860215
n ≈ 0.00556 mol
The molar mass (M) can be calculated by dividing the mass of the substance by the number of moles:
M = mass / n
Given that the flask contains 0.250 g of the vapor, we can calculate the molar mass:
M = 0.250 g / 0.00556 mol
M ≈ 44.91 g/mol
Therefore, the molar mass of the volatile liquid, to the nearest value, is approximately 44.91 g/mol.