Respuesta :
Answer: c) 100.90°C
Explanation:
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_b\times m[/tex]
[tex]\Delta T_b=T_b-T_b^0[/tex] = elevation in boiling point
i= vant hoff factor = 2 (for NaI[/tex]
[tex]NaI\rightarrow Na^++I^-[/tex]
[tex]K_b[/tex] = boiling point constant = [tex]0.512^0C/m[/tex]
m= molality
[tex]\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent i kg =48.6 g= 0.0486kg
[tex]\Delta T_b=2\times 0.512\times \frac{6.37g}{150g/mol\times 0.0486[/tex]
[tex]\Delta T_b=2\times 0.512\times \frac{6.37g}{150g/mol\times 0.0486[/tex]
[tex]\Delta T_b=0.9^0C[/tex]
[tex](T_b-1000)^0C=0.9^0C[/tex]
[tex]T_b=100.9^0C[/tex]
Thus the boiling point of the solution will be [tex]100.90^0C[/tex].
