Page 290 - Chemistry

- Compare the averaged value to the school value
If  0.1 award 1 mk
If  0.2 award ½ mk
If outside  award 0 mk

CALCULATIONS
a) Titre 1 + Titre II + Titre III = Answer
3

b) RFM of acid = 2 + 2 912) + 4(16) + 2(2 + 16)
= 126

If 500cm3 contains 6.3 g
1000cm3 contains ?
6.3 x 1000 = 12.6dm3
Concentration = 12.6g/dm3
Or 0.1 M

c) Molarity of solution C
Acid : Alkali
1 : 2

If 1000cm3 contains 0.1 moles
25cm3 contains ?

25x0.1 = 0.0025 moles
1000

From mole ratio: 25cm3 of alkali contains
0.0025 x 2 = 0.005 moles
If 25cm3 alkali contains 0.005 moles
100cm3 alkali contains 0.005x1000
25

= 0.2 moles
Molarity = 0.2 M

Procedure 2
TABLE 2
Marking should be done as in table 1

CALCULATION
a) Titre I + Titre II + Titre III = answer
3

b) 25cm3 of NaOH contains 0.005 moles
Mole ration 1 : 1
Moles of acid = 0.005 moles
If Titre in (a) of solution D contains 0.005 moles
1000cm3 of solution D contains:

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