A sample of nicotine weighing 0.3204 g was vaporized in a flask with a volume of 487 mL. This sample exrted pressure of 74.3 mm Hg at 20 degrees C. What is the molar mass of nicotine?
I know this has to do with N=P*V/R*T. and 1 atm=760 mm Hg.
Please help and show work. chemistry is not a good subject for me and i'm struggling to understand. thanks
To find the molar mass of nicotine, we can use the Ideal Gas Law equation: PV = nRT.
Given:
Pressure (P) = 74.3 mm Hg
Volume (V) = 487 mL = 0.487 L
Temperature (T) = 20 degrees C = 293 K (convert Celsius to Kelvin by adding 273)
R = 0.0821 L·atm/(mol·K) (universal gas constant)
We need to solve for n (amount of substance in moles). Rearrange the equation to solve for n:
n = PV / RT
First, let's convert the pressure from mm Hg to atm:
1 atm = 760 mm Hg
Pressure (P) = 74.3 mm Hg / 760 mm Hg/atm = 0.0975 atm
Now we substitute the values into the equation:
n = (0.0975 atm * 0.487 L) / (0.0821 L·atm/(mol·K) * 293 K)
n = 0.0470 mol
To find the molar mass (M) of nicotine, we can use the formula:
M = molar mass / number of moles
Given:
Mass of nicotine (m) = 0.3204 g
Number of moles (n) = 0.0470 mol
Rearranging the equation, we get:
M = m / n
Molar mass = 0.3204 g / 0.0470 mol ≈ 6.83 g/mol
Therefore, the molar mass of nicotine is approximately 6.83 g/mol.