Answer:
![103.06^{o}C](https://tex.z-dn.net/?f=103.06%5E%7Bo%7DC)
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:
![\Delta T_b=iK_bb](https://tex.z-dn.net/?f=%5CDelta%20T_b%3DiK_bb)
Here:
is the change in the boiling point, defined as:
![\Delta T_b=T_f - T_i](https://tex.z-dn.net/?f=%5CDelta%20T_b%3DT_f%20-%20T_i)
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:
![b=\frac{n_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7Bn_s_o_l_u_t_e%7D%7Bm_s_o_l_v_e_n_t%7D)
Therefore, we obtain:
![T_f-T_i=\frac{iK_bn_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}](https://tex.z-dn.net/?f=T_f-T_i%3D%5Cfrac%7BiK_bn_s_o_l_u_t_e%7D%7Bm_s_o_l_v_e_n_t%7D)
Solving for the final boiling point, add the initial temperature to both sides of the equation:
![T_f=T_i+\frac{iK_bn_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}](https://tex.z-dn.net/?f=T_f%3DT_i%2B%5Cfrac%7BiK_bn_s_o_l_u_t_e%7D%7Bm_s_o_l_v_e_n_t%7D)
Substitute the given variables:
![T_f=100.0^{o}C+\frac{2\cdot0.51^{o}C/m\cdot3 mol}{1 kg}= 103.06^{o}C](https://tex.z-dn.net/?f=T_f%3D100.0%5E%7Bo%7DC%2B%5Cfrac%7B2%5Ccdot0.51%5E%7Bo%7DC%2Fm%5Ccdot3%20mol%7D%7B1%20kg%7D%3D%20103.06%5E%7Bo%7DC)