stochimetery Flashcards

1
Q

Air contains 22.4% oxygen. How many moles of oxygen atoms are there in 1 liter of air at standard condition?

A

0.02
The volume of oxygen is 0.224 liters. We know that 22.4 liters contains 1 mole of oxygen molecules. Hence, 0.224 liters contains “n” moles given by
n = 0.224/22.4 = 0.01 moles of oxygen molecules
We know that one molecule of oxygen contains 2 atoms.
Therefore, number of moles of oxygen atoms is 0.02 moles

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2
Q

Non-stoichiometric compounds are formed by:

A
  • only transition elements
    Binary transition-metal compounds, such as the oxides and sulfides, are usually written with idealized stoichiometry’s, such as FeO or FeS, but these compounds are usually cation deficient and almost never contain a 1:1 cation:anion ratio. Thus a substance such as ferrous oxide is actually a nonstoichiometric compound with a range of compositions.
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3
Q

An organic compound has empirical formula C3H3O. If molar mass of the compound is 110.5 gmol-1, the molecular formula of this organic compound is (A, of C=12, H1=1.008 and O=16).

A

n = 110.5 / (12 × 3 + 1.008 × 3 + 16)
n = 110.5 / 55.02
n = 2
empirical formula = n(C3H3O)
empirical formula = 2(C3H3O)
empirical formula = C6H6O2.

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4
Q

Each molecule of Hemoglobin is made up of nearly how many atoms?

A

Hemoglobin is made up of nearly 10,000 atoms and it is 68,000 times heavier than hydrogen atom.

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5
Q

Which of the following concentration terms is/are independent of temperature?

A

molality and mole fraction

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6
Q

How many carbon atoms are present in 34.2g of sucrose (C12H22O11) Mr = 342)?

A

Number of C-atoms in sucrose
= 34.2/342 × 6×1023 × 12
= 7.2 × 1023

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7
Q

Number of electrons in 1.8ml of H2O are

A

There are total 10 electrons in 1H2O molecule
moles of H2O (n) = mass of substance / motor mass
18g of H2O = 1 mole
mass of 1.8 ml of
1.8g of water = 0.1 mole of water
Therefore, 1.8ml of H2O mean & 0.1 mole of H2O
Number of molecules = n × Avogadro number
Number of molecules = 0.1 × 6.022 × 1023 = 0.6022×1023
Number of electrons in total molecules of H2O = 0.6022×1023 × 10 = 6.022×1023

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8
Q

How many total numbers of atoms are present in 49.0g of Sulphuric acid (H2SO­4)?

A

Weight of H2SO4 = W = 49g
Molar mass of H2SO4 = M = 98g/mol
Number of moles of H2SO4 = n = W/M = 49/98 = 0.5
There are 7 atoms in the H2SO4
Total number of atoms = 7 × n × 6.022×1023 = 7 × 0.5 × 6.022×1023 = 7 × 3.02×1023

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9
Q

Which of the following ion formation is always exothermic?

A

uni negative
Due to the strong nuclear force offered by the nucleus of the neutral atom, the attraction between the nucleus and the newly added electron is very high and hence energy is released in the process of addition of electrons to the neutral atom giving rise to an exothermic process.

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10
Q

98g H2SO4 contains number of moles of ion:

A

Given amount of H2SO4 = 98g
Number of moles of H2SO4 = 98/98 = 1 mole
H2SO4 on dissociation splits up into ions such as
H2SO4 ⇌ 2H+ + SO4-2
1mole 2mole 1mole
= 2 + 1 = 3 moles of ions
Conclusion: From the equations it is clear that 1mole of H2SO4 produces 3 moles of ions.

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11
Q

weight of 1 oxygen molecule

A

One mole contains (6.02×1023 molecules) of oxygen
It weights 32g
Thus weight of 1 oxygen molecule is
32/6.023×1023 = 5.312×10-23 gm

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12
Q

The number of isotopes of elements with even mass number and even atomic number are.

A

154

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13
Q

No. of radioactive isotopes produced by artificial radioactivity

A

300

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14
Q

No. of isotopes produced naturally and idenity stable and unstable ones

A

280
240 are STABLE

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15
Q

The number of isotopes of elements with odd mass number and odd atomic number are.

A

126

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16
Q

name all the mono isotopic elements

A

5 in total
F, I, As, Au, Na

17
Q

name all the di isotopic elements

A

3
Bi , Cl, Br

18
Q

tri isotopic elements

A

5
C, H O N Ne

19
Q

tetraisotopic
pentaisotopic
hexa
nano
undeca(11)
hexdeca(16)

A

S -4
Ni -5
Ca Pd -6
Cd - 9
Sn - 11
Ag -16

20
Q

what is normality

A

the number of gram equivalent dissolved per unit volume of solvent.
formula : weight of solute/(eqv weight of solute * volume of solution)