if 95.21 g of MgCl2, a strong electrolyte is added to 650 g of water,how much would the freezing point of water be lowered? What would be freezing point of this solution?The Kf for water is 1.86 degrees C/m.

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kobenhavn kobenhavn · Answer 1 · 2019-09-03T17:09:58+02:00

Answer: Freezing point of this solution is [tex]-8.58^0C[/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 = 3 (for [tex]MgCl_2[/tex] as it dissociates to give three ions.

[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]

m= molality

Weight of solvent (water)= 650g = 0.65 kg

Molar mass of solute = 95.21 g

Molar mass of solute= 95.21 g/mol

[tex]0-T_f=3\times 1.86\times \frac{\95.21g}{95.21g/mol}\times 0.65kg[/tex]

[tex]T_f=-8.58^0C[/tex]

Thus freezing point of this solution is [tex]-8.58^0C[/tex]