Answer: 0.36 millimoles
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 75.1 cm Hg = 0.98 atm (1cmHg = 10 mm Hg)
V = Volume of gas = (51.0-42.0) ml = 9.00 ml = 0.009 L
n = number of moles = ?
R = gas constant = [tex]0.0821Latm/Kmol[/tex]
T = Temperature =[tex]27^0C=(27+273)K=300K[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{0.98atm\times 0.009}{0.0820 L atm/K mol\times 300K}=0.00036moles=0.36millimole\\ (1mole=1000millimol)[/tex]
Thus 0.36 millimoles were removed from initial air sample
The total number of mole of the gas that were removed from the initial sample is 0.36 millimole
From the question given above, the following data were obtained:
Pressure (P) = 75.1 cm Hg = 75.1 / 76 = 0.988 atm
Temperature (T) = 27 °C = 27 + 273 = 300 K
Volume (V) = 51 – 42 = 9 mL = 9 / 1000 = 0.009 L
Gas constant (R) = 0.0821 atm.L/Kmol
Using the ideal gas equation (PV = nRT), we shall determine the total number of mole of the gas that were removed from the initial sample. This is illustrated below:
0.988 × 0.009 = n × 0.0821 × 300
0.008892 = n × 24.63
Divide both side by 24.63
n = 0.008892 / 24.63
n = 0.00036 mole
Multiply by 1000 to express in millimole
n = 0.00036 × 1000
Thus, the total number of mole of the gas that were removed from the initial sample is 0.36 millimole
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