Respuesta :
Answer : The freezing point of a solution is [tex]-0.454^oC[/tex]
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]
Mass of KCl (solute) = 15 g
Mass of water (solvent) = 1650.0 g = 1.650 kg
Molar mass of KCl = 74.5 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 KCl}}{\text{Molar mass of KCl}\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 KCl 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{15g}{74.5g/mol\times 1.650kg}[/tex]
[tex]T_s=-0.454^oC[/tex]
Therefore, the freezing point of a solution is [tex]-0.454^oC[/tex]
Answer:
The freezing point of the solution containing 15 grams of KCl and 1650 grams of water is [tex]\rm -0.454^\circ\;C[/tex].
Explanation:
The freezing point of the solution can be calculated by:
Molarity = [tex]\rm \frac{mass\;of\;KCl\;}{molar\;mass\;of\;KCl\;\times\;Mass\;of\;water}[/tex]
Molarity = [tex]\rm \frac{15}{74.5;\times\;1650}[/tex]
Molarity = 1.2 [tex]\rm \times\;10^-^4[/tex] M
[tex]\rm \Delta\;T\;=\;i\;\times\;K_f\;\times\;Molarity[/tex]
[tex]\rm T^\circ\;\times\;T_f\;=\;i\;\times\;K_f\;\times\;Molarity[/tex]
[tex]\rm T_f=\;2\;\times\;1.86\;\times\;1.2\;\times\;10^-^4[/tex]
[tex]\rm T_f[/tex] = [tex]\rm -0.454^\circ\;C[/tex]
The freezing point of the solution containing KCl is [tex]\rm -0.454^\circ\;C[/tex].
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