NCERT Exemplar Solution Class 9 Science Chapter 3
NCERT Exemplar Solution Class 9 Science Chapter 3 Atoms & Molecules all Questions Answer Solution. NCERT Exemplar Class 9 Science Chapter 3 Solution PDF.
NCERT Exemplar Solution Class 9 Science Chapter 3: Overview
NCERT Exemplar Solution Class 9 Science Chapter 3 |
|
Board |
NCERT |
Topic |
Exemplar Problem Solution |
Class |
9 |
Subject |
Science |
Chapter |
3 |
Chapter Name |
Atoms & Molecules |
NCERT Exemplar Solution Class 9 Science Chapter 3 Atoms & Molecules
Multiple Choice Questions
(1) Which of the following correctly represents 360 g of water ?
(i) 2 moles of H20
(ii) 20 moles of water
(iii) 022× 1023 molecules of water
(iv) 2044 × 1025 molecules of water
a) I
b) I and IV
c) II and III
d) II and IV
Answer:- d) II and IV
Here the answer is II -20 moles of water and IV –1.2044×10 25 molecules of water
Now,
We have to find number of moles first,
Therefore,
Number of moles = mass of Water In water/Molar mass of water
Number of moles In water = 360/18
Number of moles in water= 20
Now,
Number of molecules = 20× 6.022× 10 23
= 1.2044 × 1025
Therefore, 1.2044× 10 25 molecules of water And 20 mole in 360 g of water.
2) Which of the following statements is not true about an atom ?
a) Atoms are not able to exist independently.
b) Atoms are the basic units from which molecules and ions are formed
c) Atoms are always neutral in nature.
d) Atoms aggregate in large numbers to form the matter that we can see, feel or torch
Answer:- a) Atoms are not able to exist independently.
Yes, atoms does not exist independently, atoms combine in large numbers to form an matter. These matter can be seen with the naked eyes.
3) The chemical symbol for Nitrogen gas is
a) Ni
b) N2
c) N+
d) N
Answer:- b) N2
The correct chemical symbol of the Nitrogen is N2. Because nitrogen does not exist As a molecule having two ions. thus the it has chemical formula N2. But it has chemical formula N.
4) The chemical symbol of sodium is
a) So
b) Sd
c) NA
d) Na
Answer:- d) Na
The chemical symbol of sodium is Na. Because sodium has a Latin word which is Natrium therefore it has chemical symbol Na.
(5) Which of the following would weight highest?
a) 2 mole of sucrose (C12H22O11 )
b) 2 moles of CO2
c) 2 moles of CaCO3
d) 10 moles of H2O
Answer:- c) 2 moles of CaCO3
The weight of sample= Number of moles× Molar mass of sample
Now,
a) 2 moles of C12H22O11
= 0.2× 342 = 68.4 g
b) 2 moles of CO2 = 2× 44 = 88 g
c) 2 moles of CaCO3 = 2× 100 = 200 g
d) 10 moles of H2O = 10 × 18 = 180 g
Therefore, it is clear that 2 moles of CaCO3 has the highest weight among the others.
10) A change in the physical state can be brought about
a) Only when energy is given to the system
b) Only when energy is taken out from the system
c) When energy is either given to, or taken out from the system
d) Without any energy change
Answer:- c) when energy is either given to, or taken out from the system.
Physical states can be changed with both the given in or taken out of the energy from the surrounding atmosphere. Both ways are responsible for the change in physical state.
Short Answer Questions
11) Which of the following represents a correct chemical formula? Name it.
a) CaCl
b) BiPO4
c) NaSO4
d) NaS
Answer:-
The correct chemical formula among the above given formulae is B) BiPO4 which has the name Bismuth phosphate. Here both the ions are trivalent.
12) Write the molecular formulae for the following compounds
a) Copper (II) bromide
Answer:- the molecular formula of Copper (III) bromide is CuBr2
b) Aluminium (III) nitrate
Answer:- the molecular formula of Aluminium (III) nitrate is Al(NO3)3
c) Calcium (II) phosphate
Answer:- the molecular formula of calcium (II) phosphateCa3(PO4)2
d) Iron (III) sulphide
Answer:- the molecular formula of Iron (II) sulphide is Fe2S3
e) Mercury (II) Chloride
Answer:- the molecular formula is Mercury (III) Chloride is HgCl2
f) Magnesium (II) acetate
Answer:- the molecular formula of Magnesium (II) acetate is Mg(CH3COO)2
13) Write the molecular formulae of all the compounds that can be formed by the combination of following ions
Cu2+Na+Fe3+Cl–SO 4 2- PO 43-
Answer:-
Following are some molecular formulae formed from the given compounds.
Cucl2, CuSO4, Cu3(PO4)2, Na2SO4, Na3PO4, NaCl, FeCl3, Fe2(SO4) 3, FePO4
14) Write the cations and anions present (if any) in the following compounds
a) CH3COONa
b) NaCl
c) H2
d) NH4NO3
Answer:-
Compounds
|
Anions | Cations |
a) CH3COONa
|
CH3COO- | Na+ |
b) NaCl
|
Cl- | Na |
c) H2
|
Both are cations.
It is formed a bond by sharing electrons. It is a covalent compound. |
|
d) NH4NO3
|
NO3- | NH4+ |
15) Give the formula of the compounds formed from the following sets of elements
a) Calcium and fluorine
Answer:- CaF2 – Calcium fluoride
b) Hydrogen and Sulphur
Answer:- H2S –Hydrogen sulphide
c) Nitrogen and hydrogen
Answer:- NH3- Ammonia
d) Carbon and chlorine
Answer:- CCl4–carbon tetra Chloride
e) Sodium and oxygen
Answer:-Na2O – Sodium oxide
e) Carbon and oxygen
Answer:- CO – carbon monoxide
CO2 –carbon dioxide
16) Which of the following symbols of elements are incorrect? Give their correct symbols
a) Cobalt – CO
Answer:- it is a wrong symbol
the correct symbol of cobalt is –Co
b) Carbon – c
Answer:- it is a wrong symbol
The correct symbol of Carbon is C.
c) Aluminium- AL
Answer:- it is a wrong symbol
The correct symbol of Aluminium is Al.
d) Helium – He
Answer:- it is a correctsymbol of helium.
e) Sodium – SO
Answer:- it is wrong symbol.
The correct symbol of sodium is Na.
17) Give the chemical formulae for the following compounds and complete the ratio by mass of the combining elements in each of them.
(a) Ammonia
Answer:- the chemical formula of the ammonia is NH3 .
The ratio by mass of compound is –
NH3= N:H × 3 = 14 : 1×3= 14:3
(b) Carbon monoxide
Answer:- the chemical formula of the carbon monoxide is Co.
Co= C:o = 12:16= 3:4
(c) Hydrogen chloride
Answer:- the chemical formula of the hydrogen chloride is HCl.
HCl= H:Cl = 1: 35.5
(d) Aluminium fluoride
Answer:- The chemical formula of Aluminium fluoride is AlF3.
AlF3 = Al: F×3 = 27: 19×3 =9:19
(e) Magnesium sulphide
Answer:- the chemical formula Magnesium Sulphide is MgS.
MgS = Mg: S= 24: 32= 3:4
18) State the number of atoms present in each of the following chemical species
(a) CO 3 2-
Answer:- number of atoms present in Co3 2-
Is 4.
(b) PO4 3–
Answer:- The number of atoms present in PO4 3- is 5.
(c) P2O5
Answer:- The number of atoms present in P2O5 is 7.
(d) CO
Answer:- the number of atoms present in CO is 2.
Short Answer Questions
(20) Does the solubility of a substance change with temperature? Explain with the help of an example.
Answer:- Yes, solubility of the substance is dependent upon the temperature. As we know an example for these is sugar dissolves more in hot milk than the cold milk. Thus it proved that The increase in the temperature Increases the solubility of the substances. Therefore it is temperature dependent property.
(21) Classify each of the following on the basis of their atomicity.
(a) F2
Answer:- it is diatomic.
It has 2 atoms of fluorine.
(b) NO2
Answer:- it is triatomic .
It contains two elements Nitrogen and oxygen. It contains 1 Nitrogen + 2 oxygen = 3 atoms
(c) N2O
Answer:- It is Triatomic
It has 2 atoms of nitrogen and 1 atom of oxygen thus total 3 atoms.
(d) C2H6
Answer:- it is octa Atomic.
2 atoms of carbon and 6 atoms of oxygen therefore total 8 atoms.
(e) P4
Answer:- it is tetra atomic.
A single element Phosphorus has 4 atoms.
(f) H2O
Answer:- it is triatomic.
It has 3 atoms. 2 atoms of hydrogen and 1 atom of oxygen.
(g) P4O10
Answer:- it is polyatomic.
It has 14 atoms. Phosphorus 4 atoms and 10 atoms of oxygen.
(h) O3
Answer:- it is triatomic.
Three atoms of oxygen.
(i) HCl
Answer:- it is diatomic.
It has one molecule of hydrogen and one molecule of Chlorine.
(j) CH4
Answer:- it is penta Atomic.
It has 1 atom of carbon and 4 atoms of Hydrogen. Total 5 atoms.
(k) He
Answer:- it is monoatomic.
It has 1 atom of helium.
(l) Ag
Answer:- it is monoatomic
It has one atom of Ag.
(22) You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?
Answer:- Yes. We can find that the fine powder is salt or sugar. On dissolving the powder into water, if the Solution conduct the electricity then the powder is salt. And if the Solution does not carries the electric current then the powder is sugar. Thus we can easily find that the powder is salt or sugar.
(23) Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g . Molar atomic mass of magnesium is 24g mol-1.
Answer:-
For calculating the number of moles of Magnesium,
By The formula,
Number of moles In Mg = Weight of Magnesium/ Atomic weight of Mg
= 12/24
= 0.5 moles.
Long Answer Questions
(24) Verify by calculating that
(a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
Answer:-
To find that the 5 moles of H2O and 5 moles of CO2 are same or not,
Here,
Molar mass of CO2 is =
12 + 16 × 2 = 44 g mol -1
Therefore,
molar mass of 5 moles of CO2 will be
= 44× 5= 220 g
Also,
molar mass of H2O is
= 1+1 × 16 = 18 g mol-1
Therefore
Molar mass of 5 moles of H2O is
= 18× 5 g= 90g
Therefore it is clear that the molar mass of 5 moles of H2O and 5 moles of CO2 is different.
(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5
Answer:-
Number of moles in 240g Ca metal = 240/40 = 6
Number of moles in 240 g of Mg metal = 240/24 = 10
Ratio 6:10 = 3:5
To prove that, 240 g of calcium and 240 g Magnesium elements have mole ratio of 3:5
We have to take,
For calcium
At. Weight = 40 amu
Number of moles of 240 g Ca metal = weight/Atomic Weight
= 240/6
= 40
For magnesium
At. Weight of Mg = 24 amu
Number of moles of 240 Mg metal = 240/24
10
Therefore the ratio of 240 g of Ca metal to 240 g of Mg metal is 6:10 = 3:5
25) Find the ratio by mass of the combining elements in the following compounds.
(a) CaCO3
Answer:- the ratio of mass of CaCO3 is
= Ca: Co3
=Ca: C: O×3
= 40: 12: (16×3)
= 10:3:12
(b) MgCl2
Answer:- The ratio of mass of MgCl2 is
= Mg: Cl2
= Mg : Cl× 2
= 24: (35 × 2 )
= 24 : 70
(c) H2SO4
Answer:- The ratio of mass H2SO4 is
= H2:SO4
=H× 2 : S : (O × 4)
= 1× 2 : 32 : (16×4)
= 2: 32: 64
= 1:16:32
(d) C2H5OH
Answer:- the ratio of mass of C2H5OH is
= C2:H5:O:H
=C×2:H×5:O:H
=12×2:1×5:16:1
=24:5+1:16
=24:6:16
=12:3:8
(e) NH3
Answer:- the ratio of mass of NH3 is
= N:H3
=N:H×3
=14:1×3= 14:3
(f) Ca(OH)2
Answer:- The ratio of mass of Ca(OH)2 is
= Ca:(O×H)×2
= 40:16×2:1×2
= 40:32:2
= 20:16:1
(26) Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
CaCl2(aq) → Ca 2+ (aq) + 2Cl– (aq)
Calculate the number of ions obtained from CaCl2 when 222 g of It is dissolved in water.
Answer:-
Here,
1 mole of CaCl2 = 111 g
Therefore ,
2 moles of CaCl2 = 222 g
One mole of CaCl2 gives 3 ions,
Therefore, 2 moles of CaCl2 gives = 6 ions.
No. Of ions= No. Of Moles of ions × Avogadro’s No.
= 6× 6.022×1023
= 36.132× 1023 ions
Or = 3.6132×1024 ions
(27) The difference in the mass of 100 moles each of the sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.
Answer:-
There is a difference of 1 electron between sodium atom and sodium ion.
Therefore,
For each 100 moles of sodium atom and sodium ion, there will be difference of 100 moles of electrons.
Here,
The mass of 100 moles of electrons in sodium atom and sodium ion is given,
The mass of 100 electrons = 5.48002 g
Therefore,
The mass of 1 mole of electron = 5.48002/100
Thus, mass of one electron is 9.1 × 10 -28
(28) Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass Hg and S are 200.6 g mol-1 and 32 g mol-1
Answer:-
Here, given
Molar mass of Hg = 200.6 g mol-1
Molar mass of S = 32 g mol-1
Therefore, the total of
molar mass of HgS= 200.6 + 32
= 232.6 g mol-1
Here mass of Hg in 232.6 g of HgS= 200.6 g
Therefore,
Mass of Hg in 225 g of HgS is
Thus, 194.04 g of mercury is present in the 225 g of HgS.
(29) The mass of one steel screw is 4.11 g.Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98×1024 kg) which one of the two is heavier and by how many times?
Answer:-
Here, given is
Mass of one steel screw = 4.11 g
= 4.11g
Therefore
Mass of one mole of steel screw is
=2.476× 1024g or 2.476 × 10 21 kg
Mass of earth = 5.98× 10 24. Kg
Now,
Therefore,
Earth has 2.4×103 or 2400 times more weight than the one mole of steel screw.
(30) A sample of vitamin C is known to contain 2.58× 10 24 oxygen Atoms. How many moles of oxygen atoms are present in the sample?
Answer:-
Here given is –
Total number of oxygen atoms present in vitamin C is 2.58× 10 24
1 mole of oxygen atom = 6.022 × 10 23
Therefore ,
Total 4.28 moles of oxygen are present in the sample of Vitamin C.
31) Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles od sodium atoms in another container of same weight. (a) whose container is Heavier ? (b) whose container has more number of Atoms?
Answer:-
Given,
Raunak has – 5 moles of C atoms
Krish has – 5 moles of Na atoms
Therefore,
Mass of C = 5 × 12 = 60 g
Mass of Na = 5× 23 = 115 g
Therefore Answer for a) is –Krish who has 5 moles of sodium (Na) atoms in its container is heavy than the Raunak’s Container.
Answer for b) is –both the container contain 5 moles of atoms thus both containers of Raunak and Krish has same number of atoms in it.
32) Fill in the missing data in the table 3.1
Answer:-
(33) The visible universe is estimated to contain 10 22 stars. How many moles of stars are present in the visible universe?
Answer:-
Given,
Total visible stars in universe has 10 22 stars.
Moles of stars visible in universe = ?
Therefore, there are 0.0166 moles of visible stars are present in the universe.
34) What is the SI prefix for each of the following multiples and submultiples of a unit?
(a) 103
Answer:- the SI prefix of these is Kilo.
(b) 10 -1
Answer:- the SI prefix of these is Deci.
(c) 10 -2
Answer:- the SI prefix of these is centi.
(d) 10 -6
Answer:- The SI prefix of these is micro.
(e) 10 -9
Answer:- the SI prefix of these is nano.
(f) 10-12
Answer:- The SI prefix of these is pico.
(35) Express each of the following in kilograms
(a) 84× 10 -3 mg
Answer:-= 5.84× 10 -3 × 10 -6
= 5.84 × 10 -9kg
(b) 34 g
Answer:- = 58.34 × 10 -3
= 5.834 × 10-2kg
(c) 584 g
Answer:- = 0.584 × 10-3
= 5.84 × 10 -4 kg
(d) 873× 10 -21 g
Answer:- = 5.873 × 10 -21× 10 -3
= 5.873 × 10 -24 kg
(36) Compute the difference in masses of 103 moles of magnesium atoms and magnesium ions.
(Mass of an electron = 9.1 × 10 -31 kg)
Answer:- Magnesium atom and Magnesium ion is different from each other by two electrons. I.e. Mg and Mg2+.
Therefore, 10 3 moles of Mg and Mg2+will have difference of 10 3× 2 .Now,
Mass of 103 × 2 moles of electrons
= 10 3 × 2 × 6.022 × 10 23× 9.1 × 10 -31 kg
= 2× 6.022 × 9.1 × 10 -5 kg
= 109.6004 × 10 -5 kg
= 1.096 × 10-3 kg
Thus, the difference in masses of Mg and mg2+ is 1.096 × 10 -3 kg.
(37) Which has more number of atoms?
100 g of N2 or 100 g of NH3
Answer:-
To find out which has more number of atoms,
Let’s calculate for N2
Nitrogen has two atoms present in it therefore, we have to find out number of molecules for two atoms of nitrogen.
Now, computing for NH3
NH3 has 4 atoms present in it.
= 141.69 × 10 23
Thus, it is proved that NH3 has more number of atoms as compared to N2.
38) Compute the number of ions present in 5.85 g of sodium chloride.
Answer:-
Each NaCl ( Sodium Chloride) compund has 2 ions in it. Na+ and Cl–
Now,
Total number of ions = 2
Therefore,
Total moles of ions present in 0.1 mole of NaCl is
= 0.1 × 2
= 0.2 moles
And now calculate the number of ions present in 0.2 moles of NaCl
Number of ions. = 0.2 × 6.022 × 10 23
= 1.2042 × 10 23 ions .’
5.85 g of NaCl contains total number of 1.2042× 10 23 ions .
(39) A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?
Answer:-
Here, 1 gm of gold sample contains 90% of gold
I.e. 90/100 =0.9 g of gold
Therefore,
Now,
1 mole of gold = NA atoms = 6.022× 10 23
Therefore,
0.0046 moles of gold contains
= 0.0046 × 6.022 × 10 23
= 2.77 × 10 21
Thus, 90% of gold in a gold sample will contain 2.77 × 10 21 atoms in it.
40) What are ionic and molecular compounds? Give examples.
Answer:-
Molecular Compounds:-
Molecular Compounds are the molecules which are formed by joining the atoms of same or different elements in definite proportion.
For example –Ammonia, carbon monoxide, carbon dioxide, water etc.
Ionic compounds :-
The compounds are charged species, positively charged and negatively charged. These charged species are known as ions.
Positive ion is known as cation and negatively charged ions are known as anions. When anions and cations comes together to form an compound these are known as ionic compounds.
Examples of ionic compounds –
NaCl (sodium chloride)
41) Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron 9.1× 10 -28 g). Which one is heavier?
Answer:-
Given,
Mass of electron = 9.1 × 10 -28 g.
Difference between mole of aluminium atom and mole of aluminium ion is = ?
Now, to compute the difference between them
First,
Mass of one mole
of aluminium atom = 27 g mol -1
Which is the molar mass of aluminium.
When an aluminium atom loses 3 Electrons then it becomes aluminium ion i.e. Al3+
Now,
One mole of Al3+ ion, three moles of electrons should be lost from it.
Therefore now,
Mass of 3 moles of electrons =
= 3× (9.1× 10-28) × 6.022× 10 23 g
= 164.400 × 10-5 g
= 0.00164 g
Now,
Molar mass of aluminium ion
= ( 27- 0.00164 ) g mol-1
= 26.9984 g mol-1
The difference between the moles of aluminium atom and mole of aluminium ion is
= 27 – 26.9984
= 0.0016 g
42) A silver ornament of mass ‘m’ gram is polished with gold equipment to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Answer:-
We have given here,
Mass of silver = m g
Mass of gold is 1% of silver
Now,
At. Mass of silver is 108
Number of atoms present in silver
Now, number of atoms present in gold
Gold has mass 197.
Now, calculating ratio of number of atoms present in gold to silver is
= Gold : silver
= Au : Ag
After cancellation
= 108 : 100× 197
= 108 : 19700
= 1 : 182.41
The ratio of gold to silver is 1: 182.41 .
43) A sample of ethane (C2H6) gas has the same mass as 1.5× 10 20 molecules of methane (CH4). How many C2H6 molecules does the sample of gas contain?
Answer:-
Here,
We have given that
Mass of molecules of ethane gas and methane is same 1.5 × 10 20
Mass of 1 molecule of CH4 = 16/ NA g
Now,
Mass of 1 molecule present in C2H6 = 30/ NA g
Thus,
Number of molecules present in C2H6
= 0.8 × 10 20
The number of molecules present in ethane is 0.8 × 10 20
44) Fill in the blanks
a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ________.
Answer:- In a chemical reaction, the sum of the masses of reactants and products remains unchanged. This is called Law of conservation of mass.
b) A group of atoms carrying a fixed charge on them is called _____________.
Answer:- A group of atoms carrying a fixed charge on them is called polyatomic ion.
c) The formula unit mass of Ca3(PO4)2 is ___________.
Answer:- The formula unit mass of Ca3(PO4)
Is (atomic mass of Ca×3)+( atomic mass of phosphorus ×2)+(atomic mass of oxygen × 8)
= 310
d) Formula of sodium carbonate is ___________ and that of Ammonium sulphate is _____________.
Answer:- Formula of Sodium Carbonate is Na2CO3and that of Ammonium Sulphate is (NH4)2SO4
45) Complete the following crossword puzzle (fig. 3.1) by using the name of the chemical elements. Use the data given in table 3.2.
Table 3.2
Across |
Down |
2. The element used by Rutherford during His alfa- scattering experiment. |
1. A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air. |
3. An element which forms rust on exposure to moist air. |
4. Both brass and bronze are alloys of the element. |
5. A very reactive non-metal stored under water. |
6. The metal which exists in the liquid state at room temperature. |
7. Zinc metal when treated with dilute hydrochloric acid Produces a gas of this element which when tested with burning splinter produces a pop sound. |
8. An element with symbol pb. |
Answer:-
46) (a) in this crossword puzzle (fig. 3.2), names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.
- Cl
- H
- Ar
- O
- Xe
- N
- He
- F
- Kr
- Rn
- Ne
Answer:-
(b) Identify the total number of inert gases. Their names and symbols from this cross word puzzle.
Answer:-
There are six inert gases are in the crosswords.
The name and symbols of these are –
1)Helium – He
2) Neon – Ne
3) Argon – Ar
4) Krypton – Kr
5) Xenon – Xe
6) Radon – Rn
47) Write the formulae for the following and Calculate the molecular mass for each one of them.
(a) Caustic potash
Answer:- Caustic potash has formula –KOH
Molecular mass of KOH = 36 + 16 + 1
= 56 g mol -1
(b) Baking powder
Answer:-Bakingpowder has
formula – NaHCO3
Molecular mass of
NaHCO3 = 23+1+12+3×16
= 84 g mol-1
(c) Lime stone
Answer:- formula of Lime stone is CaCO3
The molecular mass
Of CaCO3= 40 + 12+ 3×16
= 100 g mol-1
(d) Caustic soda
Answer:- The formula for Caustic soda is
NaOH
The molecular mass of NaOH = 23+16+1
=40 g mol-1
(e) Ethanol
Answer:- The molecular formula of Ethanol is C2H5OH
The molecular mass
of Ethanol is = 2×12+6×1+16
= 46 g mol-1
(f) Common salt
Answer:- the molecular formula of common salt is NaCl
The molecular mass of NaCl= 23+35
= 58 g mol-1
48) In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6H12O6. How many grams of water would be required to produce 18 g of glucose ? Compute the volume of water so consumed assuming the density of water to be 1g cm-3.
Answer:-
The reaction carried out in photosynthesis is
Now,
To form 1 mole of glucose 6 moles of water is needed as per the reaction of photosynthesis.
Therefore,
180 g of glucose needs 18× 6 = 108 g of water
Thus,
1 g of glucose will need
= Weight of glucose/Molecular weight of glucose
= 108/180 g
Now,
18 g of glucose will need
Now,
Total volume of water used
= Mass/Density
= 10.8 g/1 g cm-3
= 10.8 cm3
10.8 cm3 of water will be required to form 18 g of glucose.