Respuesta :
The approximate initial boiling point of Magnesium chloride solution has been 102.33 degree Celsius.
The pure water has boiling point 100 degree Celsius. With the addition of the solvent, the boiling point has been increased. The change in the boiling point ([tex]\Delta T_b[/tex]) has been given as:
[tex]\Delta T_b=i.K_b.b[/tex]
Where, i has been the van't Hoff factor
[tex]K_b[/tex] has been the ebullioscopic constant
b has been the molality of the sample
The dissolution of magnesium chloride in water has been given as:
[tex]\rm MgCl_2\;(aq)\;\rightarrow\;Mg^2^+\;+\;2\;Cl^-[/tex]
The i has been the amount of molecules that has been formed by the dissociation of 1 molecule. The dissociation of Magnesium chloride has been resulted in the 1 Mg and 2 Cl. Thus, the i for Magnesium chloride has been 3.
The molality (m) of the solution has been given as:
[tex]m = \dfrac{m_s_o_l_u_t_e}{mwt}\;\times\;\dfrac{1000}{m_s_o_l_v_e_n_t}[/tex]
Where, [tex]m_s_o_l_u_t_e=285\;\text {g}\\m_s_o_l_v_e_n_t=2\;\text{kg}\\mwt=95.211\;\text{g/mol}[/tex]
Substituting the values, molality of the Magnesium chloride solution (b) can be given as:
[tex]b=\dfrac{285}{95.211}\;\times\;\dfrac{1}{2}\;\text{m}\\b=1.49\;\text{m}[/tex]
The molality of the solution has been 1.49 m.
Substituting the values for change in temperature:
[tex]\Delta T_b=3\;\times\;0.52\;\times\;1.49\;^\circ \text C\\\Delta T_b=2.33\;^\circ \text C[/tex]
The rise in boiling temperature with the addition of 285 g of Magnesium chloride has been 2.33 [tex]\rm ^\circ C[/tex]. Since, the initial temperature for pure solvent has been 100 degree Celsius.
The initial temperature of Magnesium chloride solution has been the sum of two. Thus:
[tex]\text {Initial temperature}=100^\circ \text C\;+\;2.33\;^\circ \text C\\\rm Initial\;temperature=102.33\;^\circ C[/tex]
The initial temperature of magnesium chloride solution has been 102.33 degree Celsius.
For more information about the boiling point, refer to the link:
https://brainly.com/question/2153588