Question: The active mass of 64 gm of HI in a two-liter flask would be: A. 2 B. 1 C. 5 D. 0.25...

Question

The active mass of 64 gm of HI in a two-liter flask would be:
A. 2
B. 1
C. 5
D. 0.25

Answer 1

Solution

To solve this problem at first calculate the number of moles of 64 gm of HI using the mole concept. After this find out the value of active mass using the formula.

Formula used: Active mass $= \dfrac{n}{V}$

Complete step by step answer:
Before we move forward with the solution of this question, let us discuss some important concepts.
Mole: A mole is a physical quantity which represents the amount of mass of the substance required to have a collective of $6.022 \times {10^{23}}$ atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance. One mole of any substance weighs about the same as the molecular mass of that substance.
Now the number of mole(n) is the ratio of mass taken to molecular mass of HI.
Therefore, the number of moles of 64 gm of HI is,
$\dfrac{w}{M} = \dfrac{{64}}{{128}} = 0.5$
Now the formula of Active mass $= \dfrac{n}{V}$ . Where n is the number of moles and V is the volume. Therefore, the active mass of 64 gm HI is,
Active mass $= \dfrac{n}{V} = \dfrac{{0.5}}{2} = 0.25$ .
So, the correct answer is D.

Additional information:
The formula of equivalent weight is,
$Equivalent\,weight = \dfrac{{Molecular\,weight}}{{number\,of\,equivalent\,moles}}$ .
For acids the number of equivalents is equal to the number of ${H^ + }$ present in one molecule. And for bases the number of $H{O^ - }$ groups present in one molecule. The equivalent weight of any substance cannot be greater than its molecular weight. Either the value of equivalent weight is equal or less than the molecular weight of that substance.
For HI, the number of equivalents is 1.
So, $Equivalent\,weight = \dfrac{{Molecular\,weight}}{{number\,of\,equivalent\,moles}} = \dfrac{{128}}{1} = 128$

Note: Mole concept simplifies the mass relation among reactants and products such that we can base our calculation on the coefficients (numbers of molecules involved in the reaction). At the same time, mass or the quantity of substance is on lab scale in grams.