Respuesta :
Answer: The freezing point of the solution is [tex]-2.8^0C[/tex]
Explanation:-
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(0-T_f)^0C[/tex] = Depression in freezing point
i= vant hoff factor = 2 (for [tex]NaNO_3[/tex] which dissociates to give two ions )
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (water)= 200 g = 0.2 kg
Molar mass of electrolyte [tex]NaNO_3[/tex] = 85 g/mol
Mass of electrolyte added = 12.6 g
[tex](0-T_f)^0C=2\times 1.86\times \frac{12.6g}{85 g/mol\times 0.2kg}[/tex]
[tex](0-T_f)^0C=2.8[/tex]
[tex]T_f=-2.8^0C[/tex]
The freezing point of the solution is [tex]-2.8^0C[/tex]
