Olympiad Science - Atoms and Molecules
Exam Duration: 45 Mins Total Questions : 30
What is the weight of 3 gram atoms of sulphur?
- (a)
\({ 96 }_{ g }\)
- (b)
\(99_{ g }\)
- (c)
\(100_{ g }\)
- (d)
\(3_{ g }\)
The atomic mass of sulphur is 32.
1 gram atom of 5 weighs = 32 g
3 gram atoms of 5 weighs
= 3 x 32 =96 g
How many moles of oxygen atoms are present in one mole of acetic acid?
- (a)
1 mole
- (b)
3 moles
- (c)
2 moles
- (d)
6 moles
One molecule of \({ CH }_{ 3 }COOH\) contains 2 oxygen atoms.
:. One mole of \({ CH }_{ 3 }COOH\) contains 2 moles of oxygen atoms.
Calculate the weight in grams present in 0.7 moles of sodium.
- (a)
\({ 16.1 }_{ g }\)
- (b)
\({ 16.2 }_{ g }\)
- (c)
\({ 16.3 }_{ g }\)
- (d)
\(0.161_{ g }\)
1 mole of Na = 23 g
0.7 mole of Na = 23 x 0.7 = 16.1 g
How many moles are present in 540 g of glucose\({ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }\) ?
- (a)
2 moles
- (b)
3 moles
- (c)
4 moles
- (d)
1 mole
Formula of glucose = \({ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }\) Molecular weight = 180 a.m.u. One mole gram glucose = 180 g 180 g of glucose contains 1 mole of glucose 540 g of glucose contains \(\frac { 1 }{ 180g } \times 540g\)=3 moles
How much lime is obtained by burning 400g of lime stone?
- (a)
224 g
- (b)
400 g
- (c)
220 g
- (d)
320 g
\(Caco_{ 3 }\quad \underrightarrow { heat } \quad Cao+{ co }_{ 2 }(g)\)
Lime stone Lime
(40+ 12+48) (40+ 16)
= 100 g = 56g
100 g of \({ Caco }_{ 3 }\) on burning gives 56 g of
CaO.
400 g of CaC03 on burning gives = \(\frac { 56g }{ 100g } \times 400\)=224g
The approximate production of sodium carbonate per month is 424 x \({ 10 }^{ 6 }\)g while that of methyl alcohol is 320x \({ 10 }^{ 6 }\)g Which of the following is produced more in terms of number of moles?
- (a)
Sodium carbonate
- (b)
Methyl alcohol
- (c)
Both (A) and (B)
- (d)
Data insufficient
Gram formula weight or weight of 1 mole of \({ Na }_{ 2 }{ Co }_{ 3 }=106g\) Gram formula weight of methyl alcohol or weight of one mole of \({ CH }_{ 3 }{ OH }\) = 32 g
Number of moles of\({ Na }_{ 2 }{ Co }_{ 3 }\) produced per month \(=\frac { 424\times { 10 }^{ 6 } }{ 106 } =4\times { 10 }^{ 6 }\) moles.
The number of moles of \({ CH }_{ 3 }{ OH }\) produced per month = \(\\ =\frac { 320\times { 10 }^{ 6 } }{ 32 } 10\times { 10 }^{ 6 }\) moles.
Hence, methyl alcohol is produced more than sodium carbonate in terms of number of moles.
Identify the correct increasing order of molecular weights of the given compounds.
- (a)
\({ H }_{ 2 }O>{ H }_{ 2 }S>{ CO }_{ 2 }>{ SO }_{ 2 }\)
- (b)
\({ H }_{ 2 }O>{ H }_{ 2 }S<{ CO }_{ 2 }>{ SO }_{ 2 }\)
- (c)
\({ H }_{ 2 }O<{ H }_{ 2 }S<{ CO }_{ 2 }<{ SO }_{ 2 }\)
- (d)
\({ H }_{ 2 }O>{ H }_{ 2 }S>{ CO }_{ 2 }<{ SO }_{ 2 }\)
The molecular weights of \({ H }_{ 2 }O,{ H }_{ 2 }S,{ CO }_{ 2 },{ SO }_{ 2 }\) and
502 are 18,34,44 and 64 respectively. Hence,
the increasing order of molecular weights is \({ H }_{ 2 }O<{ H }_{ 2 }S<{ CO }_{ 2 }<{ SO }_{ 2 }\)
Calculate the number of moles present in 400 grams of \({ Caco }_{ 3 }\)
- (a)
2 moles
- (b)
3 moles
- (c)
1 mole
- (d)
4 moles
Formula weiqht of \({ Caco }_{ 3 }\) = 40 + 12 + 3 x 16 = 100 a.m.u.
Therefore, 1 mole of \({ Caco }_{ 3 }\) = 100 g
50,400 g is equal to 4 moles of \({ Caco }_{ 3 }\)
Calculate the number of atoms present in 6.4 g of sulphur.
- (a)
\(2.4\times { 10 }^{ 23 }\quad atoms\)
- (b)
\(2.4\times { 10 }^{ -23 }\quad atoms\)
- (c)
\(1.2\times { 10 }^{ 23 }\quad atoms\)
- (d)
\(1.2\times { 10 }^{ -23 }\quad atoms\)
Atomic weight of S = 32.06 a.m.u. 1 gram atomic weight of S = 32.06 g 6.023 x \({ 10 }^{ 23 }\)atoms of sulphur. Number of atoms in 6.4 g of S
\(=\frac { 6.4 }{ 32.06 } \times 6.023\times { 10 }^{ 23 }=1.2\times { 10 }^{ 23 }\)
How many gram atoms are present in 256g of \({ O }_{ 2 }\)
- (a)
6 gram atoms
- (b)
32 gram atoms
- (c)
14 gram atoms
- (d)
36 gram atoms
Atomic weight of an element expressed in grams is called gram atomic weight or gram atom (or) quantity of an element that contains 6.023 x\({ 10 }^{ 23 }\) atoms is called gram atom. Number of gram atoms of oxygen
\(\frac { Weight\quad of\quad element }{ Gram\quad atomic\quad element } =\frac { 256g }{ 16g } =16g\)
Calculate the number of atoms present in 71g of \({ CI }_{ 2 }\)
- (a)
\(1.205\times \)\({ 10 }^{ -24 }\)
- (b)
\(1.205\times { 10 }^{ 23 }\)
- (c)
\(1.205\times { 10 }^{ 24 }\)
- (d)
\(1.205\times { 10 }^{ -23 }\)
Numberofatoms =Numberofgramatoms x Avogadro's Number.
Atomic weight of C/ = 35.5
Number of atoms of C/ =
=\(\frac { 71.00 }{ 35.50 } \times 6.023\times { 10 }^{ 23 }\\ =2\times 6.023\times { 10 }^{ 23 }=1.205\times { 10 }^{ 24 }\)
What does the atomicity of an element tell us about?
- (a)
Physical properties
- (b)
Number of electrons
- (c)
Number of atoms in one molecule
- (d)
Its combining capacity
The atomicity of an element tells us about the number of atoms in a molecule.
Calculate the weight in grams of 0.75 gram moles of NaHC\({ O }_{ 3 }\)
- (a)
64 9
- (b)
63 g
- (c)
65 g
- (d)
66 g
G.M.W. of NaHC\({ O }_{ 3 }\)
= 23 + 1 + 12 + 16x 3 = 84 g
1 gram mole of NaHC\({ O }_{ 3 }\)weighs = 84 g
0.75 gram mole of NaHC\({ O }_{ 3 }\)weighs =0.75 x 84 = 63 g
Calculate the number of moles of helium present in 6.46 g. (The atomic weight of helium is 4 a.m.u.)
- (a)
16.15 moles
- (b)
1.615 moles
- (c)
161.5 moles
- (d)
0.1615 moles
Gram atomic weight of He = 4 g Number of moles
=\(\frac { Given\quad weight\quad of\quad element }{ Gramatomic\quad weight } =\frac { 6.46 }{ 4 } \\ =1.615\quad moles\)
Calculate the weight of 2.5 moles of \(CaC{ O }_{ 3 }\)
- (a)
200 g
- (b)
230 g
- (c)
240 g
- (d)
250 g
Formula weight of \(CaC{ O }_{ 3 }\) = 40 + 12 + 3 x 16 = 100 a.m.u. 1 mole of \(CaC{ O }_{ 3 }\)= 100 g Hence, 2.5 moles of \(CaC{ O }_{ 3 }\) == 2.5 x 100 = 250 g
How many grams of \(Na_{ 2 }C{ O }_{ 3 }\) are to be weighed to get 0.1 mole \(Na_{ 2 }C{ O }_{ 3 }\)
- (a)
10 g
- (b)
5 g
- (c)
10.6 g
- (d)
12.6 g
1 mole of\(Na_{ 2 }C{ O }_{ 3 }\) = 106 g (GMW) Weight of 0.1 mole of \(Na_{ 2 }C{ O }_{ 3 }\)
= 0.1 x 106 = 10.6 g
A molecular formula is a short form representation of 'X: Identify 'X' from the following
- (a)
A compound
- (b)
An element
- (c)
A mixture
- (d)
An alloy
A compound can be represented in a short form by molecular formula.
Calculate the number of iron atoms in a piece of iron weighing 2.8 g. (Atomic mass of iron = 56)
- (a)
\(30.11\times { 10 }^{ 23 }\) atoms
- (b)
\(30.1\times { 10 }^{ 23 }\)atoms
- (c)
\(3.0115\times { 10 }^{ 22 }\)atoms
- (d)
\(301.1\times { 10 }^{ 23 }\)atoms
1 mole of iron = gram atomic mass of iron = 56 grams
We know that 1 mole of iron element contains 6.023 x \({ 10 }^{ 23 }\)atoms of iron. 56 g of iron contains = 6.023x \({ 10 }^{ 23 }\)atoms 2.8 9 of iron contains
\(=\frac { 6.023\times { 10 }^{ 23 } }{ 56 } \times 2.8=\frac { 6.023\times { 10 }^{ 22 } }{ 2 } \\ =3.0115\times { 10 }^{ 22 }atoms\)
If the formula of chromic acid is \(H_{ 2 }{ CrO }_{ 4 }\)then what is the formula of divalent metal chromate?
- (a)
\(M{ CrO }_{ 4 }\)
- (b)
\(M_{ 2 }{ CrO }_{ 4 }\)
- (c)
\(M_{ 2 }{ (CrO }_{ 4 })_{ 3 }\)
- (d)
\(M_{ 3 }{ CrO }_{ 4 }\)
From the formula of chromic acid \(H_{ 2 }{ CrO }_{ 4 }\) the valency of chromate ion is two and the formula of divalent metal chromate is \(M{ CrO }_{ 4 }\)
Calculate the number of atoms of sulphur present in 0.5 moles of \({ Na }_{ 2 }S_{ 2 }{ O }_{ 3 }\)
- (a)
\(62.02\times 10^{ 23 }atoms\)
- (b)
\(622\times 10^{ 23 }atoms\)
- (c)
\(6.023\times 10^{ 23 }atoms\)
- (d)
\(6.02\times 10^{ -23 }\)
1mole of\({ Na }_{ 2 }S_{ 2 }{ O }_{ 3 }\) contains two gram atoms of sulphur.
0.5 mole of \({ Na }_{ 2 }S_{ 2 }{ O }_{ 3 }\)contains 1 gram atom of sulphur
1 gram atom = No. of atoms in one mole
Therefore, 0.5 mole of \({ Na }_{ 2 }S_{ 2 }{ O }_{ 3 }\)contains 6.023 x 1023atoms of sulphur.
Calculate the number of moles present in 4.9 g \({ H }_{ 2 }{ SO }_{ 4 }\)
- (a)
0.01 moles
- (b)
0.02 moles
- (c)
0.05 moles
- (d)
0.03 moles
Gram molecular weight of \({ H }_{ 2 }{ SO }_{ 4 }\) =1 x 2
+ 32 + 16x 4 = 98 g
Number of moles of \({ H }_{ 2 }{ SO }_{ 4 }\)
\(\frac { 4.9 }{ 98 } =0.05\quad moles\)
Calculate the number of moles present in 60 g of NaOH
- (a)
1.2 moles
- (b)
1.5 moles
- (c)
2.5 moles
- (d)
0.15 moles
Gram molecular weight of NaOH 23 +16+ 1 =40g
Number of moles in 60 9 of NaOH
\(\frac { 60 }{ 40 } =1.5\) moles
What weight of calcium contains the same number of atoms as those in 3 g of carbon?
- (a)
10g
- (b)
20g
- (c)
30g
- (d)
40g
No. of atoms in 3 g of carbon =0.25 x Avogadro's number.The Weight of calcium which contains the same as those in a
3 g carbon = 0.25 x GMW of Ca 0.25X40= 109
:. The number of atoms present in 3 grams(O.4 mole) of carbon is the same as in 109 (0.4 moles) of calcium.
What weight of sodium contains the same number of atoms as those in 8 grams of oxygen?
- (a)
10.5 g
- (b)
13.5 g
- (c)
11.5 g
- (d)
14.2 g
Gram atomic weight of sodium (Na) is 23 g
and that of oxygen (0) is 16 g
Number of moles of oxygen = \(\frac { Weight\quad (given) }{ GAW } =\frac { 8 }{ 16 } \)
Weight of sodium (Na) = Number of moles x GAW = 0.5 x 23 = 11.5 g
Calculate the number of moles present in 17 grams of \(AgNO_{ 3 }\)
- (a)
0.2 moles
- (b)
0.1 mole
- (c)
2 moles
- (d)
1 mole
Gram molecular weight of \(AgNO_{ 3 }\)
= (1 x 108 + 1 x 14 + 3 x 16) = 108 + 14 + 48 = 170 g
Number of moles of \(AgNO_{ 3 }\)
\(\frac { 17g }{ 170g } =\frac { 1 }{ 10 } =0.1\)moles
Identify the pair of substances having the same formula unit mass.
- (a)
Calcium chloride, potassium carbonate
- (b)
Calcium oxide, hydrochloric acid
- (c)
Carbon monoxide, ammonia
- (d)
Carbon dioxide, nitrous oxide
Carbon dioxide and nitrous oxide have the
same formula unit mass.
Option (A) \({ CaCl }_{ 2 }\) = 40 + 71 = 111
\({ K }_{ 2 }{ CO }_{ 3 }\) = (39x2) + 12 + (16X 3)
= 78 + 12 + 48 = 138
Option(B) CaO = 40 + 16 = 56
HC/ = 1 + 35.5 = 36.5
Option (C) CO = 12 + 16 = 28
NH3=14+3=17
Option (D) \({ CO }_{ 2 }\) = 12 + 16 x 2 = 44
\({ N }_{ 2 }O\) (14 x 2) + 16=44
Find the number of moles of sodium nitrate which contains 1.5 moles of oxygen atoms
- (a)
0.5
- (b)
1.5
- (c)
2.0
- (d)
1.0
1 mole of \(N{ aN }O_{ 2 }\) contains 3 moles of oxygen atoms.
:. \(N{ aN }O_{ 2 }\) containing 1.5 moles of oxygen atoms.
\(\frac { 1.5 }{ 3 } =0.5\) moles
Selenium ingestion of 90 micrograms per day causes loss of hair.How many selenium atoms are there in this sample?(Atomic weight of Se= 78.96)
- (a)
\(6.8\times 10^{ 23 }atom\)
- (b)
\(8.8\times 10^{ 17 }atoms\)
- (c)
\(8.8\times 10^{ 22 }atoms\)
- (d)
\(6.9\times 10^{ 17 }atoms\)
Atomic weight of Se = 78.96 g Number of moles in 90 micrograms
\(\frac { 90\times { 10 }^{ -6 } }{ 78.96 } =1.139\times 10^{ -6 }moles\\ 1moleofselemiumatoms=6.023\times { 10 }^{ 23 }\\ 1.139\times 10^{ -6 }\quad moles\quad ofselemium\quad atoms=1.139\times 10^{ -6 }\times 6.023\times { 10 }^{ 23 }\\ =6.9\times 10^{ 17 }atoms\)
Which of the following element's gram weight is 6 9 for 0.25 moles of that element?
- (a)
Chlorine
- (b)
Magnesium
- (c)
Boron
- (d)
Carbon
\(n=0.25moles=\frac { 1 }{ 4 } \\ moleMass(m)=6g\\ Molecular\quad weight(M)=\frac { m }{ n } \\ i.e.magnesium=\frac { 6 }{ 1/4 } =24g\)
If the formula of a chloride of a metal M is M\(C/_{ 3 }\) then what will be the formula of the phosphate of metal M?
- (a)
\(MP{ O }_{ 4 }\)
- (b)
\({ M }_{ 2 }{ PO }_{ 4 }\)
- (c)
\({ M }_{ 3 }{ PO }_{ 4 }\)
- (d)
\({ M }_{ 2 }({ PO }_{ 4 })_{ 3 }\)
\(In \ MC/_{ 3 }, \ M \ is \ trivalent \ { M }^{ +3 }(PO_{ 4 })^{ -3 }\)