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Some Basic Concepts of Chemistry

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Question
CBSEENCH12011019
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1.0 g of magnesium is burnt with 0.56 g of O2 in a closed vessel. Which reactant is left in excess and how much?

  • Mg, 0.16 g

  • O2, 0.16 g

  • Mg, 0.44 g

  • O2, 0.28 g

Solution

A.

Mg, 0.16 g

The balanced chemical equation is 
Mg + 1/2O2 --> MgO
24g    16 g        40g
From the above equation, it is clear that,
24 f Mg reacts with 16 g O2
thus, 1.0 g Mg reacts with 
16 over 24 space straight x space 0.67 space straight g space straight O subscript 2 space equals space 0.67 space straight g space straight O subscript 2 But space only space 0.56 space straight g space straight O subscript 2 space is space available space which space is space less space than space 0.67 space straight g. space Thus comma space straight O subscript 2 space si space the space limiting space reagent. further comma space 16 space straight g space straight O subscript 2 space reacts space with space Mg space 24 space straight g therefore comma space 0.56 space straight g space will space reacts space with space Mg equals space 24 over 16 space straight x space 0.56 space equals space 0.84 space straight g therefore comma space Amount space of space Mg space left space unreacted space equals space 1.0 minus 0.84 space straight g space Mg space equals space 0.16 space straight g space Mg hence space comma space Mg space is space present space in space excess space and space 0.16 space straight g space Mg space is space left space behind space unreacted.

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Question
CBSEENCH11008392
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10 g of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be


  • 2 mol

  • 3 mol

  • 4 mol

  • 1 mol

Solution

C.

4 mol

straight i space right parenthesis space straight H subscript 2 space plus 1 half straight O subscript 2 space rightwards arrow straight H subscript 2 straight O
Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount)

straight H subscript 2 space space space space space space plus space space space space space 1 half straight O subscript 2 space space space space space rightwards arrow space space space space straight H subscript 2 straight O 1 space mol space space space space space space space space space space 1 half space mol space space space space space space space space space space space space space space 1 space mol 10 over 2 space mol space space space space space space space space 64 over 31 space mol space space space space space space space space space space space ? equals space 5 mol space space space space space space space space space equals space 2 space mol space therefore comma space 1 half space mol space of space straight O subscript 2 space gives space space equals space 1 space mol space space straight H subscript 2 straight O therefore comma space 2 space mol space straight O subscript 2 space will space give space space equals space 1 space straight x space 2 space straight x space 2 space space equals space 4 space mol.

Question
CBSEENCH11008328
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20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of Mg =24)

  • 75

  • 96

  • 60

  • 84

Solution

D.

84

In the given problem we have a practical yield of MgO. For calculation of percentage yield of MgO. we need a therortical yield of MgO. For this, we shall use mole concept.
MgCO3(s) → MgO (s) + CO2(g)  .. (i)

Moles space of space MgCO subscript 3 subscript space space end subscript equals space fraction numerator weight space in space gram over denominator Molecular space weight end fraction space equals space 20 over 84 space equals space 0.238 space mol From space eq space left parenthesis straight i right parenthesis space 1 space mole space of space MgCO subscript 3 space gives space equals space 1 space mol space MgO there space 0.238 space mole space MgCO subscript 3 space will space give space equals space 0.238 space mol space MgO space equals space 0.238 space straight x space 40 space straight g space equals space 9.52 space straight g space MgO Now comma space practical space yield space of space MgO space equals space 8 space straight g space therefore comma space percent sign purity space equals space fraction numerator 8 over denominator 9.52 end fraction space straight x space 100 space equals space 84 percent sign bold Alternate bold space bold Method stack MgCO subscript 3 with 84 space straight g below space rightwards arrow space stack MgO space with 40 space straight g below plus space CO subscript 2 therefore comma space 8 space straight g space MgO space will space form space 84 over 5 straight g therefore space comma percent sign space purity space equals space 84 over 5 space straight x space 100 over 20 space equals space 84 percent sign

Question
CBSEENCH12011021
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6.02 x 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is

  • 0.02 M

  • 0.01 M

  • 0.001 M

  • 0.1 M

Solution

B.

0.01 M

Given, number of molecules of urea =6.02 x 1020
therefore ,Number of moles
=  fraction numerator 6.02 space straight x space 10 to the power of 20 over denominator straight N subscript straight A end fraction equals space fraction numerator 6.02 space x space 10 to the power of 20 over denominator 6.02 space x space space 10 to the power of 23 end fraction space equals space 1 space x 10 to the power of negative 3 end exponent space m o l v o l u m e space o f space t h e space s o l u t i o n space equals space 100 space m L space equals space 100 over 1000 L space equals space 0.1 space L C o n c e n t r a t i o n space o f space u r e a space s o l u t i o n left parenthesis i n space m o l space L to the power of negative 1 end exponent right parenthesis space equals space fraction numerator 1 space x space 10 to the power of negative 3 end exponent over denominator 0.1 end fraction space equals space m o l space L to the power of negative 1 end exponent

Question
CBSEENCH11008433
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An element, X has the following isotopic composition;


200X: 90%

199X : 8.0%

202X ; 2.0 %

The weighted average atomic mass of the naturally -occurring element X is closet to:

  • 200 amu

  • 201 amu

  • 202 amu

  • 199 amu

Solution

A.

200 amu

Weight of  200X = 0.90 x 200 = 180.00 amu
Weight of 199X  = 8.08 x 199 = 15.92 amu
Weight 0f  202X =  0.02 x 202 = 4.04 amu
Total weight = 199.06 = 200 amu

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