Respuesta :
Answer : The amount of depression will be, [tex]0.35^oC[/tex]
Explanation : Given,
Mass of sodium chloride = 5 g
Mass of water = 0.250 kg
Molar mass of sodium chloride = 58.5 g/mole
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of sodium chloride}}{\text{Molar mass of sodium chloride}\times \text{Mass of water in kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
i = Van't Hoff factor for electrolyte solution (sodium chloride) = 2
[tex]K_f[/tex] = freezing point constant = [tex]0.512^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]\Delta T_f=2\times 0.512^oC/m\times \frac{5g}{58.5g/mol\times 0.250Kg}[/tex]
[tex]\Delta T_f=0.35^oC[/tex]
Therefore, the amount of depression will be, [tex]0.35^oC[/tex]
