Answer:
0.36 g of N2.
Step-by-step explanation:
The following data were obtained from the question:
Temperature (T) = 25 °C
Volume (V) = 340 mL
Measured pressure = 730 torr
Vapour pressure = 23.76 torr
Mass of N2 =..?
First, we shall determine the true pressure of N2. This can be obtained as follow:
Measured pressure = 730 torr
Vapor pressure = 23.76 torr
True pressure =..?
True pressure = measured pressure – vapor pressure
True pressure = 730 – 23.76
True pressure = 706.24 torr.
Converting 706.24 torr to atm, we have:
760 torr = 1 atm
Therefore,
706.24 torr = 706.24 / 760 = 0.929 atm
Next, we shall convert 340 mL to L. This is illustrated below:
1000 mL = 1 L
Therefore,
340 mL = 340/1000 = 0.34 L
Next, we shall convert 25 °C to Kelvin temperature. This is illustrated below:
Temperature (K) = Temperature (°C) + 273
T(K) = T (°C) + 273
T (°C) = 25 °C
T(K) = 25 °C + 273
T (K) = 298 K
Next, we shall determine the number of mole of N2. This can be obtained as follow:
Pressure (P) = 0.929 atm
Volume (V) = 0.34 L
Temperature (T) = 298 K
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =...?
PV = nRT
0.929 x 0.34 = n x 0.0821 x 298
Divide both side by 0.0821 x 298
n = (0.929 x 0.34 ) /(0.0821 x 298)
n = 0.0129 mole
Finally, we shall determine the mass of N2 as shown below:
Mole of N2 = 0.0129 mole
Molar mass of N2 = 2x14 = 28 g/mol
Mass of N2 =.?
Mole = mass /Molar mass
0.0129 = mass of N2/ 28
Cross multiply
Mass of N2 = 0.0129 x 28
Mass of N2 = 0.36 g
Therefore, 0.36 g of N2 was collected.