Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams
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=8.30^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant =
m= molality = [tex]\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579[/tex]
[tex]8.30^0C=1\times K_f\times 0.579[/tex]
[tex]K_f=14.3^0C/m[/tex]
Let Mass of solute (KBr) = x g
[tex]8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}[/tex]
[tex]x=58.2g[/tex]
Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams
