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CHEMISTRY > The Mole Concept > Flashcards

The Mole Concept Flashcards

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1
Q

Define the term relative atomic mass (Ar)

A

the average mass of one atom of an element compared with 1/12 the mass of a carbon-12 atom

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2
Q

The mass of carbon-12 is exactly ____ units

A

12.00

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3
Q

State the Ar (relative atomic mass) of the following:

  1. Cl
  2. Na
  3. H
  4. Ca
  5. S
  6. C
A

1) 35.5
2) 23
3) 1
4) 40
5) 32
6) 12

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4
Q

The Ar/RAM is located at the ___ of the element

A

top

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5
Q

Define relative molecular mass (Mr)

A

the average mass of one molecule of a substance compared to 1/12 the mass of a carbon-12 atom

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6
Q

Mr (H2O)

A

= (2*1) +16
= 2 + 16
= 18

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7
Q

Mr (C6H1206)

A

= (612) + (121) + (6*16)
= 72 + 12+ 96
= 180

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8
Q

The relative formula is the same as relative molecular mass but applies to ____ only

A

ions

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9
Q

Define relative formula mass

A

the total of Ar of all atoms in the formula of an ionic compound

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10
Q

What is the relative formula mass of magnesium sulphate (MgSO4)?

A

Mr(MgSO4) = 24 + 32 + (4*16)
= 24 + 32 + 64
= 120

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11
Q

Define mole

A

the amount of a substance that contains 6.02 *10 ^23 particles of the substance

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12
Q

12.00 grams of carbon-12 contains _______ carbon atoms

A

6.023*10^23

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13
Q

“Avogadro number” or “Avogadro constant” is _______

A

6.023 * 10 ^23

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14
Q

When counting ‘particles’ the unit used is the _____

A

mole

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15
Q

Calculate the number of moles in 80g of Magnesium (Mg)

A

24 grams of Mg contains one mole
24= Mr
; 80 grams of Mg contains 80*1/24 = 3.33 moles

OR

n=mass/molar mass
= 80/24
= 3.33 moles

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16
Q

Calculate the no. of moles in 24g of Magnesium sulfate (MgSO4)

A

Mr(MgSO4)= 24 + 32 + (4*16)
= 120

;120 g of MgSO4 contains 1 mole
24g = 24 *1/120
= 0.2 moles

OR
n= mass/ Mr
= 24/120
= 0.2 moles

17
Q

Calculate no. moles in 18.06 *10^23 molecules of chlorine gas (Cl2)

A

6.02*10^23 molecules of gas are present in 1 mole of Cl
;18.06 * 10^23 molecules of gas are present in

18.06 *10 ^23 *1 / 6.02 *10^23
= 3 moles of Cl

18
Q

Calculate the mass of 2.6 moles of Na2CO3

A

Mr (Na2CO3) = (223) + 12 + (163)
= 106g
1 mol( Na2CO3)= 106g
;2.6 moles of Na2CO3 =

                    2.6*106/1= 275.6g

Or

mass= g*no. mol
= 106 *2.6
= 275.6 g

19
Q

Calculate the no of molecules in 2.1 moles of H2 gas

A

1 mole of H2 gas contains 6.0210^23 molecules
;2.1 moles of H2 gas contains 2.16.02*10^23/ 1
= 1.264 *10^24 molecules

20
Q

State Avogadro’s law states that

A

equal volumes of all gases at the same temperature and pressure contain the same number of molecules

21
Q

Define molar volume

A

the volume occupied by one mole of gas

22
Q

State the standard temperature and pressure (stp) in the volume of 1 mole of gas

A

22.4dm^3

23
Q

State the room temperature and pressure (rtp) in the volume of 1 mole of gas

A

24dm^3

24
Q

Calculate the volume occupied by 0.2 moles of CO2 at rtp

A

1mole of CO2 at rtp occupies 24dm^3

; 0.2 moles of CO2 at rtp occupies 24*0.2/1 = 4.8dm^3

25
Q

Calculate the number of moles in 1.3 dm^3 of NH3 at stp

A

22.4 dm3 of NH3 contains 1 mole

; 1.2 dm3 of NH3 contains 1.2*1/22.4=0.054 moles

26
Q

Calculate the volume occupied by 34g of NH3 at rtp

A

Mr NH3 = 14 + (31) = 17
; the mass of 1 mole of NH3 is 17g
17g of NH3 occupies 24 dm^3 at rtp
;34g of NH3 occupies 3424/17 =48 dm3 at rtp

27
Q

The concentration of solution can be expressed in

A

mol per dm^3 (mol dm^-3)

gram per dm^3 (g dm^-3)

28
Q

1dm^3 = _____ cm^3

A

1000

29
Q

Calculate the concentration in mol dm^-3 in 60g of NaOH in 750cm^3 of the soulution

A

Mr(NaOH) = 23 + 16 + 1 = 40
; the mass of 1 mole of NaOH is 40 g

40g of NaOH contains 1 mole
; 60g of NaOH contains 60*1/40 = 1.5 moles

750cm^3 contains 1.5 moles
; 1000cm^3 contains 1.5*1000/750 = 2.0 moles of NaOH

The concentration of the solution is therefore 2.0 mol dm^-3

30
Q

Calculate the concentration in mol dm^-3 in 80g of Na2CO3 in 800cm^3 of the soulution

A

Mr(Na2CO3) = (23 * 2) + 12 + (16*3) = 106

106g contains 1 mole of Na2CO3
; 80 g of Na2CO3 = 80*1/106 =0.755

800cm^3 contains 0.755 mols
1000cm^3 contains 1000*0.755/800= 0.943mol/dm3

The concentration of the solution therefore is 0.943mol dm^-3

31
Q

Calculate the percentage of Nitrogen present in the fertilizer Ammonium Nitrate (NH4NO3)

A

mass of 1 mole NH4NO3= 14 + (41) + 14 + (163)= 80g
Mass of Nitrogen present in 1 mole of NH4NO3= 14+14= 28g

% of Nitrogen present in NH4NO3 fertilizer
= 28/80 *100
= 35%

CHEMISTRY (1 decks)

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