0.05m NaCl is dissolved in a solution of water. Find the new freezing temperature of water ( The freezing point depression constant for water is 1.858° C/m). Please explain. Thanks!
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Hi Ganah!
This problem deals with the colligative property freezing point depression. This property basically states that when solutes are dissolved in a solvent, like water, they will cause the solvent's freezing point to lower. The key now is this equation: ΔTf = -(i)(Kf)(m) where ΔTf represents the amount of decrease of the freezing point (since the addition of solutes will lower the freezing point), Kf represents the freezing point elevation constant (we are given this in the problem), i represents the van hoff factor (the number of moles of solutes if they dissolve or not) and m represents the molality. The equation is negative on the right side because the freezing point is being lowered.
Plug in what we know into the freezing point depression equation: ΔTf = -(i)(1.858 C/M)(0.05m NaCl). We have everything except for i. The i or van hoff factor is just to account for the moles of solute actually in the solvent. Notice how NaCl actually dissolves in water( into Na+ and Cl-) so this means that in reality, there are more solutes in the solvent that we initally presumed. Because NaCl dissolves into 2 ions, i then equals 2. Now we can solve for ΔTf : ΔTf = -(2)(1.858 C/M)(0.05m NaCl) = about -0.1858 C. So to finish off the problem, knowing that water's normal freezing point is 0 degrees C right, its freezing point lowers down by 0.1858 C to -0.1858 C and that’s the new freezing temperature.
I hope this helped and feel free to ask more questions!
This problem deals with the colligative property freezing point depression. This property basically states that when solutes are dissolved in a solvent, like water, they will cause the solvent's freezing point to lower. The key now is this equation: ΔTf = -(i)(Kf)(m) where ΔTf represents the amount of decrease of the freezing point (since the addition of solutes will lower the freezing point), Kf represents the freezing point elevation constant (we are given this in the problem), i represents the van hoff factor (the number of moles of solutes if they dissolve or not) and m represents the molality. The equation is negative on the right side because the freezing point is being lowered.
Plug in what we know into the freezing point depression equation: ΔTf = -(i)(1.858 C/M)(0.05m NaCl). We have everything except for i. The i or van hoff factor is just to account for the moles of solute actually in the solvent. Notice how NaCl actually dissolves in water( into Na+ and Cl-) so this means that in reality, there are more solutes in the solvent that we initally presumed. Because NaCl dissolves into 2 ions, i then equals 2. Now we can solve for ΔTf : ΔTf = -(2)(1.858 C/M)(0.05m NaCl) = about -0.1858 C. So to finish off the problem, knowing that water's normal freezing point is 0 degrees C right, its freezing point lowers down by 0.1858 C to -0.1858 C and that’s the new freezing temperature.
I hope this helped and feel free to ask more questions!