What would the final boiling point of water be if 3 mol of nacl were added to 1 kg of water (k b =0.51^ c/(mol/kg) for water and j = 2 for nacl)? a. 98.47° c b. 101.53° c c. 96.94° c d. 103.06°c

Explanation:

The normal boiling point of water at 1 atm is . When a salt is dissolved in a solvent, in this case water, it increases the boiling point of that solvent. The final boiling point can be calculated using the boiling point elevation formula which states that:

Here:

is the change in the boiling point, defined as:

That is, the difference between the final boiling point and the initial boiling point (100 degrees Celsius);

is known as the van 't Hoff factor, in case we have a non-electrolyte/non-ionic substance, it's equal to 1, however, NaCl (aq) dissociates into 1 mole of sodium and 1 mole of chloride ions, so we have a total of 2 moles of ions per 1 mole of NaCl (aq), meaning i = 2, as the problem states;

is known as the boiling point elevation constant for the solvent;

is the molality of substance, which is found dividing moles of solute by the kilograms of solvent:

Therefore, we obtain:

Solving for the final boiling point, add the initial temperature to both sides of the equation:

Substitute the given variables:

the earth would become to overheated and everything alive on it will eventually die from overheating. plants will eventually die as well, water will evaporate, and it will be nothing but a hot planet without anything living on it.