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.

0.217 mol

Explanation:

Given:

Temperature = 14° C + 273 = 287 K.

Atmospheric pressure = 790 torr

Volume = 5 L

Vapor pressure of water = 12 torr

Now,

The vapor pressure of the Oxygen from the Law of Partial Pressures

= 790 torr - 12 torr

= 778 torr

Converting in terms of atm

we know,

1 atm = 760 torr

therefore,

778 torr = = 1.02368 atm.

also,

PV = nRT

where,

P is the pressure

V is the volume

n is number of moles

R is the gas constant = 0.0821 L-atm/mol-K

T is the temperature

on substituting the respective values, we get

1.02368 × 5 = n × 0.0821 × 287

or

n = 0.217 mol

it is a solid

explanation: