The heat of solution of ammonium chloride is 15.2 kJ/mol. If a 6.134-g sample of NH4Cl is added to 65.0 mL of water in a calorimeter at 24.5°C, what is the final temperature of the solution? The specific heat of water is 4.18 J/g·°C and the heat capacity of the calorimeter is 365 J/°C.

21.897 °C

Explanation:

Given that:

The heat of solution = 15.2 kJ/mol

Molar mass of ammonium chloride = 53.5 g/mol

So;

= 0.284 kJ/g

Also, the mass of ammonium chloride that was added to the water = 6.314 g

Then the total heat can be calculated as:

0.284 kJ/g × 6.134 g

= 1.742 kJ

= 1742 J

Volume of the water = 65.0 mL = 0.065 L

mass of the water = density × volume

= 1000 g/L × 0.065 L

= 65 g

Total mass can now be = 65 g + 6.314 g

= 71.314 g

The specific heat of water (c) = 4.18 J/g·°C

The heat capacity of the calorimeter = 365 J/°C.

Using the formula:

substituting our values and solving for ; we have:

21.897 °C

The final temperature of the solution = 21.897 °C

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."