Question

The equilibrium water vapor pressure table gives the partial pressure of water when the gas is 100.% saturated with water vapor. If the gas you collected was actually only 85% saturated with water vapor, calculate the percent error in the number of moles of N2 collected in Part I.

mass nano2 = .111g

v of wet n2 gas = 37.9 ml

partial pressure n2 = .966 atm

moles n2 collected = 1.53*10^-3

moles n2 expected = 1.61*10^-3

Answer 1

Note that total pressure = vapor pressure of water + partial pressure of N2

The volume of N2 gas collected = 37.9 ml = 37.9x 10-3 L

The partial pressure of N2 = .966 atm

Using PV= nRT, where T = 250C = 298 K and R = 0.0821 L atm mol-1K-1

n is the moles of N2 collected experimentally (caluclated value)

     (.966 atm ) ( 37.9x 10-3 L) = n x (0.0821 L atm mol-1K-1) x ( 298 K)

n = 1.50 x 10-3     .... up to 3 significant digits

So, percent error = { (moles N2 expected - moles N2 collected ) / moles N2 expected } x 100

                            = { (1.61 x 10-3 - 1.50 x 10-3 ) / (1.61 x 10-3 ) } x 100

                            = (0.11/ 1.61)x 100

                           = 6.83 %