Question

The value of GMV at STP is:
A. $22.4m{m^3}$
B. $22.4c{m^2}$
C. $22.4d{m^3}$
D. $22.4{m^3}$

Answer 1

We use the gram moles to calculate gram molecular mass.
${\text{Gram Molecular Mass = }}\dfrac{{{\text{Mass in grams}}}}{{{\text{Gram moles}}}}$
To solve this question we must know the brief about GMV (Gram Molecular Mass) and STP (Standard Temperature and Pressure).
Gram Molecular Mass is the same as the molar mass of the substance or compound but the difference is the gram molecular mass specifies the mass unit.

Complete step by step answer:
The mass of one mole of a molecular substance in grams is termed as Gram molecular mass. Gram molecular mass is considered to be the same as molar mass. The only difference is the gram molecular mass is the mass unit used. Gram molecular moles were reported in grams or grams per mole (g/mol).
To find Gram Molecular Mass:
We should use the molecular formula to calculate molar mass.
Note the relative atomic mass of each element in the formula.
Suppose ${A_x}{B_y}{C_z}$ is the molecular formula of a compound where A, B and C are the atoms; x, y and z are the numbers of atoms used respectively.
Gram Molecular Mass of ${A_x}{B_y}{C_z}$ is $(a \times x) + (b \times y) + (c \times z)g$.
To define standard conditions for temperature and which is important for the measurements of chemical and physical processes, the term commonly used is Standard temperature and pressure (STP).
The volume occupied by one gram molecular mass or one mole of gas under standard conditions (273 K and 760 mm pressure) is called gram molecular volume (GMV). Its value is 22.4 litres for each gas.
For the STP $d{m^3}$ is used thus the value is $22.4d{m^3}$ .
$n = \dfrac{V}{{22.4d{m^3}}}$
Now, the value of n is 1
$V = 22.4d{m^3}$

So, the correct answer is Option C.

Note: The reaction $KMn{O_4}$ does not characterize the color change from orange to green as the Potassium permanganate $KMn{O_4}$ solution itself is dark purple. ( refer to the hint )