Question

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

Answer 1

Answer:

$103.06^{o}C$

Explanation:

The normal boiling point of water at 1 atm is $100^{o}C$. 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:

$\Delta T_b=iK_bb$

Here:

$\Delta T_b$ is the change in the boiling point, defined as:

$\Delta T_b=T_f - T_i$

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

$i$ 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;

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

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

$b=\frac{n_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}$

Therefore, we obtain:

$T_f-T_i=\frac{iK_bn_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}$

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

$T_f=T_i+\frac{iK_bn_s_o_l_u_t_e}{m_s_o_l_v_e_n_t}$

Substitute the given variables:

$T_f=100.0^{o}C+\frac{2\cdot0.51^{o}C/m\cdot3 mol}{1 kg}= 103.06^{o}C$