Answer:
[tex]103.06^{o}C[/tex]
Explanation:
The normal boiling point of water at 1 atm is [tex]100^{o}C[/tex]. 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:
[tex]\Delta T_b=iK_bb[/tex]
Here:
[tex]\Delta T_b[/tex] is the change in the boiling point, defined as:
[tex]\Delta T_b=T_f - T_i[/tex]
That is, the difference between the final boiling point and the initial boiling point (100 degrees Celsius);
[tex]i[/tex] 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;
[tex]K_b[/tex] is known as the boiling point elevation constant for the solvent;
[tex]b[/tex] is the molality of substance, which is found dividing moles of solute by the kilograms of solvent:
[tex]b=\frac{n_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}[/tex]
Therefore, we obtain:
[tex]T_f-T_i=\frac{iK_bn_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}[/tex]
Solving for the final boiling point, add the initial temperature to both sides of the equation:
[tex]T_f=T_i+\frac{iK_bn_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}[/tex]
Substitute the given variables:
[tex]T_f=100.0^{o}C+\frac{2\cdot0.51^{o}C/m\cdot3 mol}{1 kg}= 103.06^{o}C[/tex]
