Question

5.17 grams of CaCl 2 was added to 85.0 cm 3 of water in a Styrofoam cup calorimeter. The temperature of the water changed from 21.4°C to 27.5 °C.
Calculate the molar q value (q mol) for dissolving 5.17 grams of CaCl 2 in 85.0 cm 3 of water.
Assume the heat capacity of the system is due only to the water in the system. The specific heat of water is 4.18 J/(g∙°C).

Answer 1

The amout of heat needed to change the temperature of water can be calculated as

$$Q = c{m_{{H_2}O}}\Delta T$$

where ${m_{{H_2}O}} = \rho V$ - the mass of water and $c$ is the specific heat of water. Now let's fint $Q$

$$Q = c\rho V\Delta T = 4180[\frac{{\rm{J}}}{{{\rm{kg}} \cdot {\rm{^\circ C}}}}] \cdot 1000[\frac{{{\rm{kg}}}}{{{{\rm{m}}^{\rm{3}}}}}] \cdot 85.0 \cdot {10^{ - 6}}[{{\rm{m}}^3}] \cdot (27.5[{\rm{^\circ C}}] - 21.4[{\rm{^\circ C}}]) = 2167.33[{\rm{J}}]$$

(we use that $V = 85.0[{\rm{c}}{{\rm{m}}^3}] = 85.0 \cdot {10^{ - 6}}[{{\rm{m}}^3}]$ and $c = 4.18[\frac{{\rm{J}}}{{{\rm{g}} \cdot {\rm{^\circ C}}}}] = 4180[\frac{{\rm{J}}}{{{\rm{kg}} \cdot {\rm{^\circ C}}}}]$ )

Next we can find the number of moles in $5.17[{\rm{g}}]$ grams of $CaC{l_2}$ . Calculate the molar mass of $CaC{l_2}$

$${M_{CaC{l_2}}} = {M_{Ca}} + 2{M_{Cl}} = 40.08[\frac{{\rm{g}}}{{{\rm{mol}}}}] + 2 \cdot 35.45[\frac{{\rm{g}}}{{{\rm{mol}}}}] = 110.98[\frac{{\rm{g}}}{{{\rm{mol}}}}]$$

(atomic weights of atoms can be found in the periodic table). Thus the number of moles is

$$\nu = \frac{m}{{{M_{CaC{l_2}}}}} = \frac{{5.17[{\rm{g}}]}}{{110.98[\frac{{\rm{g}}}{{{\rm{mol}}}}]}} \approx 0.047[{\rm{mol}}]$$

and the molar heat of solution is

$${C_\nu } = \frac{Q}{\nu } = \frac{{2167.33[{\rm{J}}]}}{{0.047[{\rm{mol}}]}} \approx 46113.4[\frac{{\rm{J}}}{{{\rm{mol}}}}]$$

(!!!)Note: please check the initial conditions of the problem - they can be wrong because the real value is ${C_\nu } \approx 83000[\frac{{\rm{J}}}{{{\rm{mol}}}}]$ )