Question

The weight of 112 ml of oxygen at STP, on liquefaction, would be:
A. 0.32 g
B. 0.64 g
C. 0.16 g
D. 0.96 g

Answer 1

Use the volume of gas and mass of gas at STP and determine the mass of gas of a given volume of gas. STP denotes standard temperature and pressure conditions.

Step by step answer: The standard value of temperature and pressure at STP is 273k and 1 atm respectively.

At STP volume of gas is $22400{\text{ ml}}$.

The molecular formula of oxygen gas is ${{\text{O}}_{\text{2}}}$ .

The molecular weight of ${{\text{O}}_{\text{2}}}$ is ${\text{32 g/mol}}$

So, at STP the mass of oxygen gas having volume $22400{\text{ ml}}$ is $32{\text{ g}}$.

The volume of oxygen gas given to us is ${\text{112 ml}}$.

Calculate the mass of oxygen gas having volume${\text{112 ml}}$as follows:

$112{\text{ ml }} \times \dfrac{{32{\text{ g}}}}{{22400{\text{ ml}}}} = 0.16{\text{ g}}$

So, the weight of 112 ml of oxygen at STP, on liquefaction would be 0.16 g.

Thus, the correct option is (C) 0.16 g.

Additional information: According to the ideal gas law
$PV = nRT$
Where,
P = Pressure
V = Volume
n = moles
T = temperature
R = gas constant
At STP sine temperature and pressure is constant so volume is directly proportional to moles of gas. The higher the volume of gas the higher is the moles of gas.

Note: Oxygen is diatomic gas and has a molecular formula${{\text{O}}_{\text{2}}}$. The atomic weight of O is 16g/mol and the molecular weight of ${{\text{O}}_{\text{2}}}$ is 32 g/mol. It is very important to use the correct mass of 1 mole of gas. At STP condition values of temperature, pressures are always constant. At STP 1 mole of gas has a volume of 22400 ml. Use both the volumes in the same unit.