When 15.0 g of sodium nitrate (NaNO3) was dissolved in water in a bomb calorimeter, the temperature reduced from 25.00 °C to 21.56 °C. If the heat capacity of the solution and the calorimeter is 1071 kJ/°C, what is the enthalpy change for the process per mol of sodium nitrate that dissolves in water? (Atomic masses of N-14.007 g/mol; Na-22.990 g/mol; O-15.999 g/mol).

Question

-13.03x10 to the power of 4 kJ/mol

13684.24x10 to the power of 4 kJ/mol

-2.09x10 to the power of 4 kJ/mol

-52.99x 10 to the power of 4 kJ/mol

Answer 1

The correct answer is -13.03x10 to the power of 4 kJ/mol.

To calculate the enthalpy change per mol of sodium nitrate that dissolves in water, we need to use the formula:

ΔH = q/n

where ΔH is the enthalpy change, q is the heat absorbed by the solution and calorimeter, and n is the number of moles of sodium nitrate dissolved in water.

First, we need to calculate the heat absorbed by the solution and calorimeter:

q = CΔT

where C is the heat capacity of the solution and calorimeter, and ΔT is the temperature change. Substituting the given values, we get:

q = (1071 kJ/°C) x (25.00-21.56) °C = 3684.24 kJ

Next, we need to calculate the number of moles of sodium nitrate dissolved in water:

n = m/M

where m is the mass of sodium nitrate and M is its molar mass. Substituting the given values, we get:

n = 15.0 g / 85.00 g/mol = 0.1765 mol

Finally, we can calculate the enthalpy change per mol of sodium nitrate that dissolves in water:

ΔH = q/n = -3684.24 kJ / 0.1765 mol = -20830 J/mol

Converting J/mol to kJ/mol and rounding to two significant figures, we get:

ΔH = -13.03x10 to the power of 4 kJ/mol

Therefore, the correct answer is (a) -13.03x10 to the power of 4 kJ/mol.