Page 285 - Chemistry

Complete table – 1 mk
Decimal - 1 mk
Accuracy - 1 mk
Principle of averaging – 1 mk
Final Answer - 1 mk

3
b) Average volume of solution M used V 1 = (20.0 + 20.0) cm
2
3
= 20.0 cm
c) Mass per litre = 23.5 √½ = 0.0599 √½
Molar mass 392

d) 25 x Answer (c) = 25 x 0.0599 √½
1000 1000
= 0.0014987 √½

3
-
e) 20 cm of solution M contains Answer in (d) x 1 moles of MnO 4
5
= 0.0014987 x 1√½
5
= 0.0002997 moles. √½

3
∴ 1000 cm of solution M contains 1000 x Answer in (d)
20 5
= 1000 x 0.0002997 √½
20
= 0.014985 moles √½
f) Table II

I II III
3
Final burette reading (cm ) 19.4 38.8. 19.4
3
Initial burette reading (cm ) 0.0 19.4 0.0
3
Volume of solution M used (cm ) 19.4 19.4 19.4

Complete table – 1 mk
Decimal - 1 mk
Accuracy - 1 mk
Principle of averaging – 1 mk
Final Answer - 1 mk

g) Average volume of solution M used, V 2 = (19.4 + 19.4 + 19.4) cm 3
3
3
= 19.4 cm

h) Average volume x Answer in (e)
1000
19.4 x 0.014985√½ = 0.0002907 √½
1000
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