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When 10 g of anhydrous CaCl_(3) is dissolved in 200 g of water, the temperature of the solution rises by 7.7 ^(@) C. Calculate the heat of hydration of CaCl_(2) to CaCl_(2). 6 H_(2)O. Given that the heat of dissolution of CaCl_(2). 6 H_(2)O is 19.08 kJ mol^(-1) . Assume specific heat of the solution to be same as that of water, i.e., 4.184 J g^(-1) K^(-1) |
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Answer» Solution :Step I. Calculationof heat of dissolution of anhydrous `CaCl_(2)` Molar mass of `CaCl_(2) = 40 + 2 xx 35.5 = 111g mol^(-1)` Heat EVOLVED in dissolved of 10g`CaCl_(2)` in 200 G WATER `=m xx c xx Detla t` `= 200 xx 4.184 xx 7.7J` `:. `Heat of dissolutionper mole `=( 200 xx 4.184 xx 7.7) /( 10 ) xx 111 J = 71.52kJ` i.e,`Delta_("disso.")H ( CaCl_(2)) = - 71.52kJ mol^(-1)` Step II. Calculation of heat of hydration . Aim `: CaCl_(2)(s) + 6H_(2)O(l)rarr CaCl_(2).6 H_(2)O(s)` We have `:` `Delta_("disso.") H(CaCl_(2))= 71.52kJ mol^(-1) ` ( Calculated above) `Delta_("disso.") H ( CaCl_(2). 6 H_(2)O) = 19.08 kJ mol^(-1) ` ( Given ) i.e.,`(i)CaCl_(2) (s) + aq rarr CaCl_(2)(aq), DeltaH_(1) = - 71.52kJ mol^(-1)` (ii) `CaCl_(2). 6 H_(2)O(s) + aq rarr CaCl_(2)(aq), DeltaH_(2) = 19.08kJ mol^(-1)` Reaction (i) can be written in two steps as `:` `CaCl_(2)(s) + 6 H_(2)O(l) rarrCaCl_(2). 6 H_(2)O(s) , Delta H - Delta H _(3)` `CaCl_(2). 6 H_(2) O (s) + aq rarr CaCl_(2)(aq), DeltaH = Delta H _(4)` `:. Delta H _(3) + Delta H _(4)DeltaH _( 1) = - 71.52 kJ mol^(-1)` But `DeltaH_(4) = DetlaH _(2) = 19.08 kJ mol^(-1)` `:. DeltaH _(3) + 19.08 -71.52kJ mol^(-1)` or `Delta H _(3) = - 90.6 kJ mol^(-1)` |
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