Respuesta :
Answer : The temperature at which water freeze will be, [tex]-12.7^oC[/tex]
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]
Mass of NaCl (solute) = 100 g
Mass of water (solvent) = 500 g = 0.500 kg
Molar mass of NaCl = 58.5 g/mole
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}\times \text{Mass of water in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = ?
[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]
i = Van't Hoff factor = 2 (for NaCl electrolyte)
[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]0^oC-T_s=2\times (1.86^oC/m)\times \frac{100g}{58.5g/mol\times 0.500kg}[/tex]
[tex]T_s=-12.7^oC[/tex]
Therefore, the temperature at which water freeze will be, [tex]-12.7^oC[/tex]
