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

103.06 C

Explanation: Ape

Answer 2

`103.06^(o)C`

Step-by-step 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=(n_s_o_l_u_t_e)/(m_s_o_l_v_e_n_t)`

Therefore, we obtain:

`T_f-T_i=(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+(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+(2\cdot0.51^(o)C/m\cdot3 mol)/(1 kg)= 103.06^(o)C`