NEET Exam  >  NEET Tests  >  Chemistry Class 11  >  RC Mukherjee Test: Mole Concept - NEET MCQ

RC Mukherjee Test: Mole Concept - NEET MCQ


Test Description

14 Questions MCQ Test Chemistry Class 11 - RC Mukherjee Test: Mole Concept

RC Mukherjee Test: Mole Concept for NEET 2024 is part of Chemistry Class 11 preparation. The RC Mukherjee Test: Mole Concept questions and answers have been prepared according to the NEET exam syllabus.The RC Mukherjee Test: Mole Concept MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RC Mukherjee Test: Mole Concept below.
Solutions of RC Mukherjee Test: Mole Concept questions in English are available as part of our Chemistry Class 11 for NEET & RC Mukherjee Test: Mole Concept solutions in Hindi for Chemistry Class 11 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt RC Mukherjee Test: Mole Concept | 14 questions in 15 minutes | Mock test for NEET preparation | Free important questions MCQ to study Chemistry Class 11 for NEET Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
RC Mukherjee Test: Mole Concept - Question 1

Vapour density of a gas is 22. Its molecular mass will be:​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 1

Molecular mass = 2 x vapour density
= 2 x 22
= 44
Therefore, molecular mass of a gas will be 44.

RC Mukherjee Test: Mole Concept - Question 2

What is the mass of 0.20 mole of C2H5OH (ethanol)?

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 2

Molecular Mass of Ethanol
= 2(12) + 5 (1) + 16 + 1 
= 24 + 5 + 17 
= 46
Now,
Moles (n) = Given Mass (m)/Molecular Mass (M)
Thus, to = nM
= (0.2) (46)
= 9.2

RC Mukherjee Test: Mole Concept - Question 3

Consider the reaction between hydrogen and oxygen gases to form water. Which of the following is/are not conserved in the reaction?
2H2(g) + O2(g) → 2H2O(l)​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 3

In the equation both side of reaction moles of atoms is conserved ie, both reactant and product contain 4H and 2O but there is 3 (2+1) moles of reactant molecules turns into 2 moles of product molecules. 

RC Mukherjee Test: Mole Concept - Question 4

Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O2(g) → C12H22O11 (s)

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 4

C = 84/12 = 7 mole

H2 = 12 g = 6 mole

O2 = 56/22.4 = 5/2 mole

12C + 11H2 + 11/2 O2 → C12H22O11

L.R. = O2

11/2 mole O2 produce 1 mole sucrose 5/2 mole O2 will for 5/11 mole sucrose mass of sucrose = 5/11 × (mol. mass)

= 5/11 × 342

= 155.45 g

RC Mukherjee Test: Mole Concept - Question 5

Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 5

Let atomic weight of x = Mx

atomic weight of y = My

we know,

mole = weight /atomic weight

a/c to question,

mole of xy2 = 0.1

so,

0.1 = 10g/( Mx +2My)

Mx + 2My = 100g -------(1)

for x3y2 ; mole of x3y2 = 0.05

0.05 = 9/( 3Mx + 2My )

3Mx + 2My = 9/0.05 = 9× 20 = 180 g ---(2)

solve eqns (1) and (2)

2Mx = 80

Mx = 40g/mol

and My = 30g/mole

RC Mukherjee Test: Mole Concept - Question 6

Number of atoms present in 52 g of helium is:
(NA is Avogadro’s constant Gram atomic mass of He is 4 g)​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 6

Gram atomic mass of helium is 4 g so number of atoms present in 52 g of helium
52/4 × NA = 13 NA

RC Mukherjee Test: Mole Concept - Question 7

The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would be

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 7

As vapour density of mixture is 25,

Molar mass of mixture is 2×25=50.

Now, assume mole % of B in mixture is x.

RC Mukherjee Test: Mole Concept - Question 8

Atomic mass of Cl is 35.5 g. Calculate the mass of 4.50 moles of chlorine gas, Cl2.​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 8

Mass of 1 mole of Clis 71 g. Hence mass of 4.50 moles is 71 X 4.50 = 319.5 g

RC Mukherjee Test: Mole Concept - Question 9

Atomic mass of bromine is 80 g. The mass of four moles of molecular bromine (Br2) is:​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 9

Mass of one mole of Br2  is 160g. Hence mass of four moles of molecular bromine is 4 X 160 = 640 g.

RC Mukherjee Test: Mole Concept - Question 10

Number of nitrogen atoms present in 1.4 g of N2

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 10

Given weight of N₂ gas = 1.4 g

Molar mass of N₂ gas = 28 g

So, mole = given mass/ molar mass

⇒ mole = 1.4/28 = 1/20 mole

Now, number of molecules = mole × avogadro number

⇒ number of molecules = 1/20 × 6.022 × 10²³

⇒ number of molecules = 3.011 × 10²²

Now, we are asked for number of atoms. In N₂, there are 2 atoms, so to obtain number of atoms we will multiply with 2 in number of molecules.

⇒ number of atoms = 2 × 3.011 × 10²²

⇒ number of atoms = 6.022 × 10²²

RC Mukherjee Test: Mole Concept - Question 11

Volume of 17g of NH3 at N.T.P. will be​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 11

Molar mass of NHis 17g. So the volume of 17g of NH3 i.e 1 mole NH3 at N.T.P. will be 22.4L.

RC Mukherjee Test: Mole Concept - Question 12

Which of the following has maximum number of moles?​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 12

1g hydrogen has the maximum number of moles because atomic mass of hydrogen is smallest among all.

RC Mukherjee Test: Mole Concept - Question 13

Molar mass of F2 is 38 g. How many atoms are present in 0.147 mole of F2?​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 13

One molecule of F2  contains 2 atoms of fluorine.

1 mol of fluorine contains 2 X 6.023 X 1023 atoms

So, 0.147 moles will contain 0.147 X 2 X 6.023 X 1023 = 1.76 X 1023 atoms of flourine.

RC Mukherjee Test: Mole Concept - Question 14

Mass of 6.022 x 1022 molecules of CO2 is about:​

Detailed Solution for RC Mukherjee Test: Mole Concept - Question 14

1 mole of a substance is equal to its molecular mass expressed in terms of grams.
Gram molecular mass of CO2 = 12 + (16 Χ 2) = 12 + 32 = 44 g
Mass of 6.023 X1022 molecules of CO2 is 44g/10 = 4.4g.

172 videos|309 docs|152 tests
Information about RC Mukherjee Test: Mole Concept Page
In this test you can find the Exam questions for RC Mukherjee Test: Mole Concept solved & explained in the simplest way possible. Besides giving Questions and answers for RC Mukherjee Test: Mole Concept, EduRev gives you an ample number of Online tests for practice

Up next

172 videos|309 docs|152 tests
Download as PDF

Up next

How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!