Page 13 - Focus SPM KSSM F4 2020 - Chemistry

Chemistry Form 4 Chapter 3 The Mole Concept, Chemical Formula and Equation
SPM Highlights
EXAMPLE 3.13
What is the mass of magnesium powder that 63.5 g of gas X occupies 5600 cm at STP. What is
3
has 3.612 × 10 magnesium atoms? the molar mass of gas X?
24
-1
3
[Relative atomic mass: Mg, 24. Avogadro’s [Molar volume: 22.4 dm mol at STP]

A 56

constant, NA: 6.02 × 10 mol ] B 127
-1
23
Solution C 180
Number of moles of Mg = number of particles ÷ N A D 254
3.612 × 10 24 Examiner’s tip
=
6.02 × 10 23 First, analyse the question. The question gives the
= 6 mol mass and volume of the gas. To determine the molar
Therefore, the mass of magnesium powder mass of gas X, students need to calculate the mass
3
of 1 mole of gas X or 22 400 cm at STP.
= number of moles × molar mass Number of moles of gas X that occupies 5600 cm 3
= 6 × 24 volume of gas
= 144 g = molar volume Chapter
5 600
=
22 400
EXAMPLE 3.14 = 0.25 mol 3
So, the mass of 0.25 mole of gas X is 63.5 g.
What is the volume of 100 g of ammonia gas, This means that the mass of 1 mole of gas X
NH at STP? 63.5
3
[Relative atomic mass: H, 1; H, 14. Molar = 0.25
volume: 22.4 dm mol at STP] = 254 g
-1
3
So, the molar mass of gas X is 254.
Solution Answer: D
Number of moles of NH = mass ÷ molar mass
3 100
= Checkpoint 3.2
14 + 3(1) [Relative atomic mass: H,1; N, 14; O,16; Ne, 20; Na,
100 23; S, 32; K, 39; Fe, 56; Cu, 64; Br, 80. Avogadro’s
= 17 = 5.88 mol constant, N : 6.02 × 10 mol . Molar volume: 24 dm
3
-1
23
So, volume of NH mol at room conditions.]
A
-1
3
= number of moles × molar volume Q1 A sample consists of 2.5 moles of carbon dioxide
= 5.88 × 22.4 gas, CO .
2
= 131.71 dm (a) How many molecules are there in the sample?
3
(b) Hence, calculate the number of atoms in
the sample.
(c) What is the volume of the sample at room
EXAMPLE 3.15
conditions?
How many hydrogen molecules, H are there in Q2 Calculate the number of moles of each of the
2
6 dm of the gas at room conditions? following.
3
(a) 13.8 g of sodium, Na
[Molar volume: 24 dm mol at room (b) 40.4 g of potassium nitrate, KNO
-1
3
3
conditions. Avogadro’s constant, N : 6.02 × Q3 An experiment requires 0.5 mole of copper(II)
A
-1
23
10 mol ] sulphate, CuSO .
4
(a) Calculate the molar mass of copper(II)
Solution sulphate.
Number of moles of H = volume ÷ molar volume (b) What is the mass of copper(II) sulphate
2 6 required in the experiment?
= 24 = 0.25 mol Q4 What is the mass of carbon that has twice the
amount of atoms found in 11.2 g of iron?
So, the number of H molecules Q5 Arrange the following samples of gases in an
2
= number of moles × Avogadro’s constant ascending order of volume.
= 0.25 × 6.02 × 10 23 X: 5 g of neon gas, Ne
= 1.505 × 10 molecules Y: 10 g of ammonia gas, NH 3
23
Z: 16 g of bromine gas, Br
2
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03 SPM CHEMISTRY F4.indd 41 27/02/2020 11:23 AM