Question

What would be the change in the temperature of boiling if 55 g of nacl is added to 774 of water

Answer 1

The boiling point elevation can be calculated using the following Clausius-Clapeyron and Raoult laws:

$\Delta T_b = i·K_b·b_B$ ,

where $\Delta T_b$ is the difference of the boiling point of the solution and the boiling point of the solvent (water), $i$ is the Van't Hoff factor, accounting for the dissociation (for NaCl, which is a strong electrolyte, $i=1.9$ ), $K_b$ is the ebullioscopic constant that is specific for the solvent (for water, its value equals 0.512 °C·kg/mol) and $b_B$ is the molality of the solute (NaCl).

The molality of the number of the moles of the solute per kilogram of the solvent:

$b_{NaCl} = \frac{n_{NaCl}}{m_{water}} = \frac{m_{NaCl}}{m_{water}·M_{NaCl}} = \frac{55}{774·10^{-3}·58.44} = 1.216$ mol/kg.

In this equation, the molar mass $M_{NaCl}$ is used to calculate the number of the moles from the mass of NaCl and equals 58.44 g/mol. Therefore, the unit of mass of NaCl is kept in g, but the mass of the solvent (water) is converted to kg, dividing by a factor of 1000.

Finally, the boiling point elevation is:

$\Delta T_b = 1.9·0.512·1.216 = 1.18$ °C.

Answer: when 55 g of nacl is added to 774 g of water, the boiling point elevation is 1.18 °C, or 1.2 °C.