Question:

A gaseous mixture of three gases A. B and C has a pressure of 10 atm. The total number of moles of all the gases is 10. If the partial pressures of Aand B are 3.0 and 1.0 atm respectively and if C has a mol/wt. of 2.0. What is the weight of C in g present in the mixture ?

Updated On: Jul 5, 2022

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The Correct Option is C

Solution and Explanation

Total pressure of mixture of gases A, B and C = 10 atm. Number of total moles = 10 Partial pressure of A = $\frac{No. of \, moles \, of \, A}{10} \times 10 $ $\therefore$ 3 = No. of moles of A $\because$ Partial pressure of B = $\frac{No.of \, moles \, of \, B}{10} \times 10$ = No. of moles of B. $\therefore$ No. of moles of C = 10 - (3 +1) = 6. Now 1 mole of C = 2g $\therefore$ 6 mol of C = 2 x 6 = 12 g

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There are three States of Matter:

The three states of matter are as follows:

Solid State:

The solid-state is one of the fundamental states of matter.
Solids differ from liquids and gases by the characteristic of rigidity.
The molecules of solids are tightly packed because of strong intermolecular forces; they only oscillate about their mean positions.

Liquid State:

The molecules in a liquid are closely packed due to weak intermolecular forces.
These forces are weaker than solids but stronger than that of gases.
There is much space in between the molecules of liquids which makes their flowing ability easy.

Gaseous State:

In this state of matter, distances between the molecules are large (intermolecular distance is in the range of 10^-7-10^-5 cm.
The intermolecular forces experienced between them are negligible.
Thus, translatory, rotatory and vibratory motions are observed prominently in gases.