1.1 g of CaCI2 dissolved in 50 cm3 of water caused a rise in temperature of 3.4°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)
-
-71.1
- -4.18
- +71.1
- +111.0
The correct answer is: A
Explanation
Q = mct = (50 x 4.18 x 3.4) / (1000)
= 0.71606 kj
1.1 g of \(CaCl_2\) contains
(1.1) / (111) = 0.01 moles of \(CaCl_2\)
If 0.01 liberates 0.7106 quantity of heat.
1 will liberate (0.7106 x 1) / (0.01)
Exothermic = 71.06 = -71.1