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
jamb 2011chemistry

The correct answer is B. 0.20mole

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.

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