From the diagram above, find the amount of solute deposited when 200 cm(^3) of the solution is cooled from 55°C to 40°C.
-
A.
0.10 mole -
B.
0.20mole -
C.
0.01 mole -
D.
0.02 mole
Correct Answer: Option B
Explanation
From the diagram,
55°C = 7 moles; 40°C = 6 moles.
Amount of solute deposited = 7 – 6 = 1 mole.
1000 cm(^3) = 1 mole
200 cm(^3) = x
x = (frac{200 times 1}{1000})
= 0.20 mole.