Question

Given, 600 mL of ozonized oxygen at STP was found to weigh one gram. What is the volume of ozone in the ozonized oxygen?
(A)- 200 mL
(B)- 150 mL
(C)- 100 mL
(D)- 50 mL

Answer 1

Hint: 22400 mL of ozone and oxygen at STP will weigh 48 g and 32 g respectively.

Complete answer:
-Let us assume the volume of ${{O}_{3}}$ being x mL.
-Now, the volume of ${{O}_{2}}$ the present will be = (600-x) mL.
-22400 mL of ${{O}_{3}}$at STP will weigh= $16\times 3=48g$
 22400 mL of ${{O}_{2}}$ at STP will weigh= $16\times 2=32g$
-Let us now calculate the weight of ${{O}_{3}}$ and ${{O}_{2}}$ ,
The weight of x ml of ${{O}_{3}}=\dfrac{x\times 48}{22400}g$ .
The weight of (600-x) mL of ${{O}_{2}}=\dfrac{(600-x)}{22400}\times 32$
Therefore the total weight of ozonized ${{O}_{2}}$ (600 mL) $=\dfrac{48x}{22400}+\dfrac{(600-x)\times 32}{22400}=1.0$

Thus, the value of x=200 mL. i,e. Option (A)

Additional information:
-Standard temperature and pressure (STP) are the standard sets of conditions for experimental measurements that allow us to make comparisons between different sets of data.
-IUPAC defines standard temperature and pressure (STP) as ‘a temperature of 273.15 K (\[0{}^\circ ,32{}^\circ \] F) and an absolute pressure of exactly ${{10}^{5}}$ Pa (100 kPa, 1 bar).
-These conditions are the most commonly used conditions to define the volume term in a Normal cubic meter ($N{{m}^{3}}$ ).

Note:
-Normal Temperature and Pressure are defined as air or gas at $20{}^\circ C$(293.15 K,$68{}^\circ F$ ).
-NPT also is known as the volume occupied by 1 mole of gas at NTP is 22.4L and it is a standard condition for testing and documentation of experiments.
-You should not get confused between STP and SATP. Standard Ambient Temperature and Pressure is a reference with a temperature of $25{}^\circ C$ (298.15 K) and a pressure of 101.325 kPa.