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,...

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

Answer Details

This is a problem in thermodynamics that involves the calculation of the heat of reaction (∆H) for the dissolution of CaCl2 in water. The equation for the dissolution reaction is: CaCl2 (s) + H2O (l) → Ca2+ (aq) + 2Cl- (aq) The given information is: mass of CaCl2 = 1.1 g, volume of water = 50 cm3, temperature rise = 3.4°C, specific heat of water = 4.18 J K-1 g-1, and atomic weights of Ca and Cl are 40 and 35.5, respectively. To calculate ∆H, we need to use the equation: ∆H = -q / n where q is the heat absorbed by the solution (water + CaCl2), and n is the number of moles of CaCl2 dissolved. First, we need to calculate the number of moles of CaCl2: n = mass / molar mass n = 1.1 g / (40 g/mol + 2 x 35.5 g/mol) = 0.010 mol Next, we need to calculate the heat absorbed by the solution: q = m x c x ΔT where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the temperature rise. The mass of the solution can be calculated from the density of water: density = mass / volume mass = density x volume mass = 1 g/cm3 x 50 cm3 = 50 g Now we can calculate q: q = 50 g x 4.18 J K-1 g-1 x 3.4°C = 714 J Finally, we can calculate ∆H: ∆H = -714 J / 0.010 mol = -71.4 kJ/mol Therefore, the answer is -71.1 kJ, which is closest to.