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Test: Percentage Composition (NCERT) - NEET MCQ


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5 Questions MCQ Test Chemistry Class 11 - Test: Percentage Composition (NCERT)

Test: Percentage Composition (NCERT) for NEET 2024 is part of Chemistry Class 11 preparation. The Test: Percentage Composition (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Percentage Composition (NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Percentage Composition (NCERT) below.
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Test: Percentage Composition (NCERT) - Question 1

0.48 g of a sample of a compound containing boron and oxygen contains 0.192 g of boron and 0.288 gof oxygen. What will be the percentage composition of the compound?

Detailed Solution for Test: Percentage Composition (NCERT) - Question 1

Mass of compound taken = 0.48 g
Mass of boron in sample = 0.192 g
Mass of oxygen in sample = 0.288 g
Percentage of B = 0.192/0.48 x 100 = 40%
Percentage of O = 0.288/0.48 x 100 = 60%

Test: Percentage Composition (NCERT) - Question 2

A compound contains two elements 'x' and 'Y' in the ratio of 50% each. Atomic mass of 'x' is 20 and 'Y' is 40. What can be its simplest formula?

Detailed Solution for Test: Percentage Composition (NCERT) - Question 2

Percentage of element 'X' = 50;
Mole ratio = 50/20 = 2.5
Percentage of element 'Y' = 50, mole ratio = 50/40 = 1.25
For 'X' simple ratio = 2.5/1.25 = 2
For 'Y' simple ratio = 1.25/1.25 = 1
Formula = X2Y

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Test: Percentage Composition (NCERT) - Question 3

The empirical formula of a compound is CH2O2.What could be its molecular formula?

Detailed Solution for Test: Percentage Composition (NCERT) - Question 3

Since empirical formula is multiplied by n to get molecular formula.
CH2O2 will give only C2H4O4 as its molecular formula. (CH2O2)n where n = 1, 2, 3,... etc.

Test: Percentage Composition (NCERT) - Question 4

A gas has molecular formula (CH)n. If vapour density of the gas is 39, what should be the formula of the compound?

Detailed Solution for Test: Percentage Composition (NCERT) - Question 4

Mol. wt. = 2 x V.D. = 2 x 39 = 78
(CH)n = (13)n or 13 x n = 78
n = 78/13 = 6
∴ Molecular formula = C6H6

Test: Percentage Composition (NCERT) - Question 5

Choose the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively and its molecular mass is 160.

Detailed Solution for Test: Percentage Composition (NCERT) - Question 5

For element Fe, mole of atoms = 69.9/56 = 1.25
For element O, Mole of atoms = 30.1/16 = 1.88
Mole ratio of Fe = 1.25/1.25 = 1
Mole ratio of O = 1.88/1.25 = 1.5
Simplest whole number ratio of Fe and O = 2, 3
Empirical formula of compoimd = Fe2O3
Molecular mass of Fe2O3 = 160

Molecular formula = Fe2O3

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