NEET | Chemistry | Some Basic Concept of Chemistry | Question with explanation | Previous Year Questions

Chemistry

Some Basic Concepts of Chemistry

1. The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is

(a) 40 (b) 10

(NEET 2019)

Ans - (d)

2. The density of 2 M aqueous solution of NaOH is 1.28 g/cm³ . The molality of the solution is (Given that molecular mass of NaOH = 40 g mol^-1]

(a) 1.20 m (b) 1.56 m

(Odisha NEET 2019)

Ans. (c)

3. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄ The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

(a) 1.4 (b) 3.0

(NEET 2018)

Ans – (c)

4. In which case is number of molecules of water maximum?

(a) 18 mL of water

(b) 0.18 g of water

(NEET 2018)

Ans – (a)

5. Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂weighs 9g, the atomic weights of X and Y are

(a) 40,30

(b) 60, 40

(d) 30, 20

(NEET-II 2016)

Ans. (a)

6. What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO, is mixed with 50 mL of 5.8% NaCl solution? (Ag=107.8, N=14,O=16, Na = 23, Cl = 35.5)

(a) 3.5g (b) 7g

(2015)

Ans – (b)

7. If Avogadro number NĄ, is changed from 6.022 x 10²³ mol^-1 to 6.022 x 10²⁰ mol^-1 this would change

(a) the mass of one mole of carbon

(b) the ratio of chemical species to each other in a balanced equation

(d) the definition of mass in units of grams.

(2015)

Ans. (a)

8. The number of water molecules is maximum in

(a) 1.8 gram of water

(b) 18 gram of water

(d) 18 molecules of water.

(2015)

Ans – (c)

9. A mixture of gases contains H₂ and O₂ gases in the ratio of 1:4 (w/w). What is the molar ratio of the two gases in the mixture?

(a) 16:1 (b) 2:1

(2015, Cancelled)

Ans – (d)

10. Equal masses of H, O, and methane have been taken in a container of volume V at temperature 27°C in identical conditions. The ratio of the volumes of gases H,:0, : methane would be

(a) 8:16:1

(b) 16:8:1

(d) 8:1:2

(2014)

Ans – (d)

11. When 22.4 litres of H_2(g) is mixed with 11.2 litres of Cl_2(g) , each at S.T.P, the moles of HCl_(g) formed is equal to

(a) 1 mol of HCl_(g)

(b) 2 mol of HCl_(g)

(d) 1.5 mol of HCl_(g)

(2014)

Ans – (a)

12. 1.0 g of magnesium is burnt with 0.56 g O₂ in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24,0 = 16)

(a) Mg, 0.16 g

(b) 0, 0.16 g

(d) 0, 0.28 g

(2014)

Ans – (a)

13. 6.02 x 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of solution is (a) 0.001 M

(b) 0.1 M

(d) 0.01 M

(NEET 2013)

Ans – (d)

14. In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO₃ which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine)?

(a) X₂Cl₂

(b) XCl₂

(d) X₂Cl

(Karnataka NEET 2013)

Ans – (b)

15. Which has the maximum number of molecules among the following?

(a) 44 g CO₂

(b) 48 g O₃

(d) 64 g SO₂

(Mains 2011)

Ans – (c)

16. The number of atoms in 0.1 mol of a triatomic gas is (N_A = 6.02 x 10²³ mol^-1 )

(a) 6.026 x 10²²

(b) 1.806 x 10²³

(d) 1.800 x 10²²

(2010)

Ans – (b)

17. 25.3 g of sodium carbonate, Na₂CO₃ is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na⁺ and carbonate ions, CO₃^2- respectively

(Molar mass of Na₂CO₃= 106 g mol^-1)

(a) 0.955 M and 1.910 M

(b) 1.910 M and 0.955 M

(d) 0.477 M and 0.477 M

(2010)

Ans – (b)

18. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

(a) 3 mol

(b) 4 mol

(d) 2 mol

(2009)

Ans – (b)

19. What volume of oxygen gas (O₂) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane gas (C₃H₈) measured under the same conditions?

(a) 5L (b) 10 L

(2008)

Ans – (a)

20. How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g HCl?

(a) 0.011 (b) 0.029

(2008)

Ans – (b)

21. An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be

(a) CHO (b) CH₄O

(2008)

Ans – (c)

22. An element, X has the following isotopic composition : ²⁰⁰X : 90% ⁹⁹X: 8.0% ²⁰²X : 2.0% The weighted average atomic mass of the naturally occurring element X is closest to

(a) 201 amu (b) 202 amu

(2007)

Ans – (d)

23. The maximum number of molecules is present in

(a) 15 L of H₂ gas at STP

(b) 5 L of N₂ gas at STP

(d) 10 g of O₂ gas.

(2004)

Ans – (a)

24. Which has maximum molecules?

(a) 7 g N₂

(b) 2 g H₂

(d) 16 g O₂

(2002)

Ans – (b)

25. Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxidase anhydrous enzyme is

(a) 1.568 x 10⁴

(b) 1.568 x 10³

(d) 2.136 x 10⁴

(2001)

Ans – (a)

26. Molarity of liquid HCl, if density of solution is 1.17 g/cc is

(a) 36.5 (b) 18.25

(2001)

Ans- (c)

27. Specific volume of cylindrical virus particle is 6.02 x 10^-2 cc/g whose radius and length are 7 Å and 10 Å respectively. If N_A = 6.02 x 10²³, find molecular weight of virus.

(a) 15.4 kg/mol

(b) 1.54 x 10⁴ kg/mol

(d) 3.08 x 10²kg/mol

(2001)

Ans – (a)

28. Volume of CO₂ obtained by the complete decomposition of 9.85 g of BaCO₃ is

(a) 2.24 L (b) 1.12 L

Ans – (b)

29. Oxidation numbers of A, B, C are +2, +5 and -2 respectively. Possible formula of compound is

(a) A₂(BC)₂

(b) A₃(BC₄)₂

(d) A₃(B₂C)₂

(2000)

Ans – (b)

30. The number of atoms in 4.25 g of NH₃ is approximately

(a) 4 x 10²³

(b) 2 x 10²³

(d) 6 x 10²³

(1999)

Ans – (d)

31. Given the numbers : 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers is

(a) 3, 3 and 4 respectively

(b) 3, 4 and 4 respectively

(d) 3, 3 and 3 respectively.

(1998)

Ans – (d)

32. Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of haemoglobin is

(a) 4 (b) 6

(1998)

33. In the reaction,

4NH_3(g)+ 5O_2(g)→ 4NO_(g) + 6H₂O_(l)

when 1 mole of ammonia and 1 mole of O₂ are made to react to completion :

(a) all the oxygen will be consumed

(b) 1.0 mole of NO will be produced

(d) all the ammonia will be consumed.

(1998)

Ans – (a)

34. 0.24 g of a volatile gas, upon vaporisation, gives 45 mL vapour at NTP. What will be the vapour density of the substance? (Density of H, = 0.089 g/L)

(a) 95.93 (b) 59.93

(1996)

Ans - (b)

35. The amount of zinc required to produce 224 mL of H₂ at STP on treatment with dilute H₂SO₄ will be

(a) 65 g

(b) 0.065 g

(d) 6.5 g

(1996)

Ans – (c)

36. The dimensions of pressure are the same as that of

(a) force per unit volume

(b) energy per unit volume (c) force

(d) energy

(1995)

Ans – (b)

37. The number of moles of oxygen in one litre of air containing 21% oxygen by volume, under standard conditions, is

(a) 0.0093 mol

(b) 2.10 mol

(d) 0.21 mol.

(1995)

Ans – (a)

38. The total number of valence electrons in 4.2 g of Nion is (NA is the Avogadro's number)

(a) 2.1 NA

(b) 4.2 NA

(d) 3.2 N.

(1994)

Ans – (c)

39. A 5 molar solution of H₂SO₄ is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be

(a) IN (b) 0.1 N

(1991)

Ans – (a)

40. The number of gram molecules of oxygen in 6.02 x 10²⁴ CO molecules is

(a) 10 g molecules

(b) 5 g molecules

(d) 0.5 g molecules.

(1990)

Ans – (b)

41. Boron has two stable isotopes, ¹⁰B(19%) and ¹¹B(81%). Calculate average at. wt. of boron in the periodic table.

(a) 10.8 (b) 10.2

(1990)

Ans – (a)

42. The molecular weight of o, and so, are 32 and 64 respectively. At 15°C and 150 mm Hg pressure, one litre of O₂ 'N’ molecules. The number of mol two litres of So, under the same co of temperature and pressure will

(a) N/2 (b) N

(1990)

Ans – (c)

43. A metal oxide has the formula Z₂O₃. It can be reduced by hydrogen to give free and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is (a) 27.9 (b) 159.6

(1989)

Ans – (d)

44. Ratio of C_p and C_v of a gas 'X' is number of atoms of the gas 'X' pr 11.2 litres of it at NTP will be

(a) 6.02 x 10²³

(b) 1.2 x 10²³

(d) 2.01 x 10²³

(1989)

Ans – (d)

45. What is the weight of oxygen for the complete combustion of ethylene?

(a) 2.8 kg (b) 6.4 kg

(1989)

Ans – (c)

46. The number of oxygen atoms in 4.4 g of CO₂ is

(a) 1.2 x 10²³

(b) 6 x 10²²

(d) 12 x 10²³

(1989)

Ans – (a)

47. At S.T.P. the density of CCI, vapour in g/L will be nearest to

(a) 6.87 (b) 3.42

(1988)

Ans (a)

48. One litre hard water contains 12.00 mg Mg²⁺Milli-equivalents of washing soda required to remove its hardness is

(a) 1

(b) 12.16

(d) 12.16 x 10^-3

(1988)

Ans – (a)

49. 1 cc N₂O at NTP contains

(a) (1.8/22.4) x10²² atoms

(b) (6.02/22400) x 10²³ molecules

(d) all of the above.

(1988)

Ans – (d)

Explanation

1. (d)

Haber's process,

N₂ + 3H₂ → 2NH₂

2 moles of NH₂ are formed by 3 moles of H₂

20 moles of NH₃ will be formed by 30 moles of H.

2. (c)

Density = 1.28 g/cc,

Conc. of solution = 2 M

Molar mass of NaOH = 40 g mol^-1

Volume of solution = 1 L = 1000 mL

Mass of solution = d x V

= 1280 g

Mass of solute = n x Molar mass = 2 x 40 = 80 g

Mass of solvent = (1280 – 80) g = 1200 g

Number of moles of solute

= (80/40) = 2

Molality = (2 x 1000)/1200

= 1.67 m

3. (c)

H₂O absorbed by H₂SO₄ Gaseous mixture (containing CO and CO₂) when passed through KOH pellets, CO₂gets absorbed.

Moles of CO left (unabsorbed) =

Mass of CO=moles x molar mass =

4. (a)

Mass of water = Vxd=18x1 = 18 g

Molecules of water = mole x N_A = (18/18) N_A= N_A

(b)

Molecules of water = mole x N_A= (0.18/18) N_A=10^-2 N_A

(c)

Moles of water

= (0.00224/22.4) = 10^-4

Molecules of water = mole x N_A= 10^-4 NA

(d)

Molecules of water = mole x N_A= 10^-2 N_A

5. (a)

Let atomic weight of element X is x and that of element Y is y. For XY₂, n = w/Mol. wt.

0.1 = 10/(x + 2y)

x + 2y = (10/0.1) = 100 …(i)

For X₃Y₂, n = w/Mol. wt.

0.05 = 9/(3x + 2y)

3x + 2y = (9/0.05) = 180 …(ii)

On solving equations (i) and (ii), we get y = 30

x + 2(30) = 100

x= 100 - 60 = 40

6. (b)

16.9% solution of AgNO₃ means 16.9 g of AgNO₃ in 100 mL of solution. 16.9 g of AgNO₃ in 100 mL solution

= 8.45 g of AgNO₃ in 50 mL solution.

Similarly, 5.8 g of NaCl in 100 mL solution = 2.9 g of NaCl in 50 mL solution.

The reaction can be represented as:

Mass of AgCl precipitated

= 0.049 x 143.3

= 7.02 = 7 g

7. (a)

Mass of 1 mol (6.022 x 10²³atoms) of carbon = 12 g

If Avogadro number is changed to 6.022 x 10²⁰atoms then mass of 1 mol of carbon = (12 x 6.022 x 10²⁰)/

(6.022 x 10²³) = 12 x 10^-3g

8. (c)

1.8 gram of water = (6.023 x 10²³ x 1.8) / (18)

= 6.023 x 10²² molecules

18 gram of water = 6.023 x 10²³ molecules 18 moles of water

= 18 x 6.023 x 10²³molecules

9. (d)

Number of moles of H₂ = (1/2)

Number of moles of O₂ = (4/32)

Hence, molar ratio

= (1/2) : (4/32) = 4 : 1

10. (c)

According to Avogadro's hypothesis, ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, no. of moles

= Mass/Mol. Mass

So, the ratio is

11. (a)

1 mole = 22.4 litres at S.T.P.

Here, Cl₂ is limiting reagent. So, 1 mole of HCl_(g) is formed.

12. (a)

Here, O₂ is limiting reagent.

Mass of Mg left in excess = 0.0066 x 24 = 0.16 g

13. (d)

Moles of urea

= (6.02 x 10²⁰)/(6.02 x 10²³)

=0.001

Concentration of solution

= (0.001 x 100) x 1000

= 0.01 M

14. (b)

Millimoles of solution of chloride

= 0.05 x 10 = 0.5

Millimoles of AgNO₃ solution

= 10 x 0.1 = 1

So, the millimoles of AgNO₃are double than the chloride solution.

XCl₂ + 2AgNO₃ → 2AgCl + X(NO₃)₂

15. (c)

8 g H₂ has 4 moles while the others has 1 mole each.

16. (b)

No. of atoms

= N_A x No. of moles x 3

= 6.023 x 10²³ x 0.1 x 3

= 1.806 x 10²³

17. (b)

Given that molar mass of

Na₂CO₃ = 106 g

Molarity of solution

= (25.3 x 1000)/(106 x 250)

= 0.955 M

Na₂CO₃→ 2Na⁺ + CO₃^2-

[Na⁺] = 2 [Na₂CO3)

= 2 x 0.955 = 1.910 M

[CO₃^2-] = [Na₂CO₃] = 0.955 M

18. (b)

H₂ + (1/2) O₂ →H₂O

2g 16g 18 g

1 mol 0.5 mol 1 mol

10 g of H₂ = 5 mol and

64 g of O₂ = 2 mol

In this reaction, oxygen is the limiting reagent so amount of H₂O produced depends on that of O₂

Since 0.5 mol of O₂ gives 1 mol H₂O

2 mol of O₂ will give 4 mol H₂O

19. (a)

C₃H_g + 5O₂ → 3CO₂ + 4H₂O

1 vol. 5 vol. 3 vol. 4 vol.

According to the above equation 1 vol. or 1 litre of propane requires to 5 vol. or 5 litre of O₂ to burn completely.

20. (b)

Formation of moles of lead (II) chloride depends upon the no. of moles of PbO which acts as a limiting factor here. So, no. of moles of PbCl₂ formed will be equal to the no. of moles of PbO i.e. 0.029.

21. (c)

22. (d)

Hence, empirical formula of the compound would be CH₃O.

Average isotopic mass of X

= (200 x 90+199 8+202 x 2)/

(90+8+2)

= (18000+1592 +404)/(100)

=199.96 a.m.u.

= 200 a.m.u. 100

23. (a)

24. (b)

Number of molecules = moles x N_A

Molecules of N₂

= (7/14) N_A = 0.5 N_A

Molecules of H₂ = 2N_A Molecules of NO₂ = 16/46 = 0.35 N_A

Molecules of O₂ = 0.5 N_A

2Gh₂ (1g mole H₂) contains maximum molecules.

25. (a)

In peroxidase anhydrous enzyme 0.5% Se is present means, 0.5 g Se is present in 100 g of enzyme. In a molecule of enzyme one Se atom must be present. Hence, 78.4 g Se will be present in

(100/0.5) x 7.84 = 1.568 x 10⁴

26. (c)

Density = 1.17 g/cc.

1 cc. solution contains 1.17 g of HCI

Molarity

= (1.17 x 1000)/(36.5 x 1)

= 32.05

27. (a)

Specific volume (vol. of 1 g) of cylindrical virus particle

= 6.02 x 10^-2 cc/g

Radius of virus, r = 7 Å

= 7 x 10^-8 cm

Volume of virus = πr²l

= (22/7) x (7x10^-8)² x10x10^-8

= 154 x 10^-23 cc

wt. of one virus particle

=Volume/ Specific volume

= (154 x10^-23)/ (6.02 x10^-2) g

Molecular wt. of virus

= wt. of N_A particles

= (154 x10^-23 x 6.02 x 10^-23) (6.02x10^-2)g/mol

= 15400 g/mol = 15.4 kg/mol

28. (b)

BaCO₃ → BaO + CO₂

197.34 g 22.4 L at N.T.P.

9.85 g (22.4/197.34) x 9.85

= 1.118

= 9.85 g BaCO₃ will produce 1.118 L CO₂ at N.T.P. on the complete decomposition.

29. (b)

In A₃(BC₄)₂,

(+2) x 3 + 2[+5 +4(-2)]

= +6 + 2 (-3) = 0

Hence, in the compound A₃(BC₄)₂, the oxidation no. of ‘A’, 'B' and ‘C’ are +2, +5 and -2 respectively.

30. (d)

No. of molecules in 4.25 g NH₃ = (4.25/17) x 6.023 x 10²³

= 2.5 X 6.023 x 10²²

Number of atoms in 4.25 g NH₃ = 4 x 2.5 x 6.023 x 10²²

= 6.023 x 10²³

31. (d)

Zeros placed left to the number are never significant, therefore the no. of significant figures for the numbers.

161 cm = 0.161 cm and 0.0161 cm are same, i.e., 3

32. (a)

Quantity of iron in one molecule

= (67200/100) x 0.334

= 224.45 atm

No. of iron atoms in one molecule of haemoglobin

= (224.45 / 56) = 4

33. (a)

4NH_3(g)+5O_2(g) → 4NO_(g) + 6H₂O_(l)

4 mole 5 mole 4 mole 6 mole

= 1 mole of NH₃requires = 5/4

= 1.25 mole of oxygen while 1 mole of O₂ requires =4/5

= 0.8 mole of NH₃

Therefore, all oxygen will be consumed.

34. (b)

Weight of gas = 0.24 g, Volume of gas = 45 mL

= 0.045 litre and density of H₂ = 0.089

weight of 45 mL of H₂

= density x volume

= 0.089 x 0.045

= 4.005 x 10 g

Therefore, vapour density

= (Weight of certain volume of substance)/ (Weight of same volume of hydrogen)

= 0.24 / (4.005x10^-3)

=59.93

35. (c)

Zn + H₂SO₄ → ZnSO₄+H₂

65 22400 ml

Since 65 g of zinc reacts to liberate 22400 mL of H₂ at STP, therefore amount of zinc needed to produce 224 mL of H₂ at STP

= (65/22400) x 224 = 0.65 g

36. (b)

Pressure= Force/Area

Therefore, dimensions of pressure = MLT^-2/L²

= ML^-1T^-2

and dimensions of energy per unit volume = Energy/volume

= ML² T^-2/L³ = ML^-1T^-2

37. (a)

Volume of oxygen in one litre of air = (21/100) X 1000

= 210 mL

Therefore, no. of mol

= (210/22400) = 0.0093 mol

38. (c)

Each nitrogen atom has 5 valence electrons, therefore total number of electrons in N₃^- ion is 16. Since the molecular mass of N₃ is 42, therefore total number of electrons in 4.2 g of N₃^- ion

= (4.2/42) x 16 x N_A

=1.6 N_A

39. (a)

5 M H₂SO₄ = 10 N H₂SO₄

N₁V₁ = N₂V₂

=10 × 1 = N₂ x 10

= N₂ = 1 N

40. (b)

Avogadro's No.

N_A = 6.02 x 10²³ molecules 6.02 x 10²⁴ CO molecules

= 10 moles CO

= 10 g atoms of O

= 5 g molecules of O₂

41. (a)

Average atomic mass

= (19x10+81x11)/100

= 10.81

42. (c)

If 1 L of one gas contains N molecules, 2 L of any gas under the same conditions will contain 2 N molecules.

43. (d)

Z₂O₃ + 3H₂ → 2Z + 3H₂O Valency of metal in Z₂O₃ = 3 0.1596 g of Z₂O₃ react with 6 mg of H₂

[1 mg = 0.001 g = 10^-3 g]

1 g of H₂ react with = (0.15965/0.006)

= 26.6 g of Z₂O₃

Eq. wt. of Z₂O₃ = 26.6

Now, Eq. wt. of Z + Eq. wt. of O

= Eq. wt. of Z + 8 = 26.6

= Eq. wt. of Z= 26.6 - 8

= 18.6

At. wt. of Z = 18.6 x 3 = 55.8

[Eq. wt. = Atomic wt./Valency of metal]

44. (a)

Here, C_P /C_y = 1.4, which shows that the gas is diatomic.

22.4 L at NTP

= 6.02 x 10²³molecules

11.2 L at NTP = 3.01 x 10²³molecules

Since gas is diatomic.

11.2 L at NTP = 6.02 x 10²³ atom

45. (c)

C₂H₄ +3O₂ → 2CO₂ + 2H₂O

28 g 96 g

For complete combustion

2.8 kg of C₂H₄ requires

= (96g/28 g) x2.8 kg of O₂

= (96/28) x2.8x10³g

= 9.6 x 10³g = 9.6 kg of O₂

46. (a)

1 mol of CO₂ = 44 g of CO₂:

4.4 g CO₂ = 0.1 mol CO₂

= 6 x 10²² molecules

[Since, 1 mole CO₂ = 6 x 10²³molecules]

= 2 x 6 x 10²² atoms of O

= 1.2 x 10²³ atoms of O

47. (a)

Weight of 1 mol CCl₄ vapour

= 12 + 4 x 35.5 = 154 g Density of CCl₄ vapour

= (154/22.4) gL^-1

= 6.875 gL^-1

48. (a)

Mg²⁺ + Na₂CO₃ → MgCO₃ + 2Na⁺

1g eq. of Mg²⁺ = 12g of Mg²⁺ = 12000 mg

Now, 1000 millieq. of

Na₂ CO₃ = 12000 mg of Mg²⁺ .

1 millieq. of Na₂ CO₃ = 12 mg of Mg²⁺

49. (d)

As we know,

22400 cc of N₂O contain 6.02 x 10²³ molecules .

1 cc of N₂O contain 6.02 x 10²³ molecules

Since in N₂O molecule there are 3 atoms

1cc N₂O = (3x6.02 x 10²³)/22400 atoms

= (1.8 x 10²²)/224 atoms

No. of electrons in a molecule of

N₂O = 7 + 7 + 8 = 22

Hence, no. of electrons

= (6.02 x 10²³/22400) x 22 electrons

= (1.32/224) x 10²³electrons

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