Question

Calculate the amount of oxygen gas collected by the displacement of water at 14◦C if the atmospheric pressure is 790 Torr and the volume is 5 L. The vapor pressure of water at 14◦C is 12 Torr. 1. 0.217 mol 2. 0.00335 mol 3. 0.0184 mol 4. 4.46 mol 5. 0.224 mol

Answer 1

Answer : The amount of oxygen gas collected are, 0.217 mol

Explanation :

Using ideal gas equation :

$PV=nRT$

where,

P = pressure of gas = $(790-12)torr=778torr=1.02atm$     (1 atm = 760 torr)

V = volume of gas = 5 L

T = temperature of gas = $14^oC=273+14=287K$

n = number of moles of gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1.02atm)\times (5L)=n\times (0.0821L.atm/mol.K)\times (287K)$

$n=0.217mole$

Thus, the amount of oxygen gas collected are, 0.217 mol