Question

The density ammonia at \[\text{30}{}^\circ \text{C}\] and 5 atm pressure is?
A. $2.42\text{g}{{\text{L}}^{-1}}$
B. $\text{1}\text{.71g}{{\text{L}}^{-1}}$
C. $3.42\text{g}{{\text{L}}^{-1}}$
D. $\text{3}\text{.84g}{{\text{L}}^{-1}}$

Answer 1

Hint: For this problem, we have to use the Ideal gas equation which is PV = nRT here, P is considered as the pressure of the gas, V as the volume of gas, n is the number of moles of gas whereas R is the gas constant which is fixed and T is the temperature by evaluating this equation we will get density formula.

Complete Step-by-step answer:
- In the question, we have to calculate the density of ammonia at the given temperature and pressure.
- As we know that Ideal gas equation follows the gas law and is given by PV = nRT here, P is considered as the pressure of a gas, V as the volume of gas, n is the number of moles of gas whereas R is the gas constant and T is the temperature.
- Also, we know that the density of the substance is given by the ratio of mass to the volume of the molecule.
- It is given in the question that the temperature is \[\text{30}{}^\circ \text{C}\] or 303 K and the pressure is 5 atm.
- We can$\text{P = }\dfrac{\text{nRT}}{\text{V}}$ write the ideal equation as:
- Also, we know that the number of moles is given by the ratio of mass to the molecular weight of the substance i.e.
$\text{n = }\dfrac{\text{mass}}{\text{Molecular weight}}$
$\text{P = }\dfrac{\text{Mass }\times \text{ RT}}{\text{Molecular weight }\times \text{ V}}$ …… (1)
- Equation (1) can also be written as
$\text{d = }\dfrac{\text{P}\times \text{ Molecular weight}}{\text{RT}}$ …… (2)
Now, putting all the values in equation (2) we will get:
$\text{d = }\dfrac{5\times \text{ 17}}{0.821\text{ }\times \text{ 303}}\text{ = 3}\text{.42g}{{\text{L}}^{1-}}$

Therefore, option C is the correct answer.

Note: In the given question, the molecular weight of the ammonia i.e. $\text{N}{{\text{H}}_{3}}$will be $\text{14 }\times \text{ (3 }\times \text{ 1) = 17}$. And when the unit of pressure is given in 'atm' then the value of a gas constant is taken as $0.821\,\text{L atm }{{\text{K}}^{1-}}\text{ mo}{{\text{l}}^{1-}}$.