1.1 g of CaCI2 dissolved in 50 cm3 of water caused a rise in temperature of3.4°C. The heat of reaction, ∆H for CaCI2 in kj per mole is?

(Ca = 40, CI = 35.5, specific heat of water is 4.18 jk-1

  • A -71.1
  • B -4.18
  • C +71.1
  • D +111.0
jamb 1991chemistry

The correct answer is A. -71.1

Q = mct = (50 x 4.18 x 3.4) / (1000)

= 0.71606 kj

1.1 g of call2 contains

(1.1) / (111) = 0.01 moles of call2

If 0.01 liberates 0.7106 RS

1 will liberate (0.7106 x 1) / (0.01)

Exothermic = 71.06 = -71.1

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