Page 235 - Chemistry

100
= 720g/l  ½
KClO 3 = 39 + 35.5 + 3 x 16 = 122.5
molarity = 720g/l  ½
122.5gmol -1
 ½
= 5.878mol/l
o
(iv) Mass dissolved at 62 = 116g
o
Mass dissolved at 42 = 66g  ½ 1
 ½
mass crystallized out = 50g

 ½
(b) (i) (25 x 0.2M) = 0.005mol
1000
(ii) 0.005mol (mole ration Acid: Base = 1:1)
3
(iii) 20cm contain 0.005mol  ½
3
3
25cm contain = (250cm x 0.005mol)
3
20cm  ½
= 0.0625mol

-1
(iv) Mass = (0.0625x 4ogmol ) = 2.5g  1

(v) Mass of solvent = 28g – 2.5g = 25.5g  1  1
solubility = (100 x 2.5)
 ½
25.5
= 9.804g/100g water  ½
15. a) Solubility refers to the maximum mass of solute dissolving in a 100g of a solvent at a
particular temperature

b) i) Fractional crystallization
ii) Scale = 1 mk
Plotting = 1 mk
Curve L = 1 mk
Curve M = 1 mk
iii) I = Actual value from students curve + 1C
II = Actual value from students curve + 1

iv) Mass per litre = 1000 X Actual value in iii (II)
100
Concentration = Above answer
132
= _________ M
+
-
16. (a) (i) Conductivity decreases wince H ions form he acid are neutralized by OH ions
from the base. This reduces the concentration of ions available for conductivity.
-
(ii) Conductivity increases since the OH ions accumulate after complete neutralization of the
-
acid OH increases conductivity.
(iii) Neutralization leads to the formation of a slat. The ions in the salt are responsible for
conducting of electricity.
+
(iv) They yield different concentration of H ions
+
For HNO 3 – dissociates completely hence more H ions
+
HCOOH – dissociates partially hence less H ions

(b) 2HCOOH (aq) + Na 2CO 3(aq) 2HCOONa (aq) + H 2O (l) + CO 2(g)
moles of HCOOH = 50 x 0.1

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