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Question: Calculate the amount of \(CaC{l_2}\) (molar mass = 111 \(gmo{l^{ - 1}}\)) which must be added to 580...

Calculate the amount of CaCl2CaC{l_2} (molar mass = 111 gmol−1gmo{l^{ - 1}}) which must be added to 580 g of water so that freezing point lower by 2K, assuming CaCl2CaC{l_2} is completely dissociated:
[Kf{K_f} For water = 1.86KKgmol−1 KKgmo{l^{ - 1}}].

Explanation

Solution

This question is asked from the topic of colligative properties of solutions. Whenever we add a solute particle in a solvent, then the freezing point of the solution will decrease in a certain amount. This lowering of the freezing point is one of the colligative properties.

Complete step by step answer:
We know that the colligative properties formulas are useful for finding out the molecular mass of the solute particles.
The formula for the decrease in freezing point is, ΔTf=i×(Kf×WB×1000)WA×MB\Delta {T_f} = i \times \dfrac{{\left( {{K_f} \times {W_B} \times 1000} \right)}}{{{W_A} \times {M_B}}}
Where ii is the van’t hoff factor, Kf{K_f} is cryoscopic constant, WB{W_B} is the mass of solute particles, WA{W_A} is the mass of solvent and MB{M_B} is the molar mass of solute.
In the question, it is given that:
Kf{K_f} for water = 1.86 KKgmol−1KKgmo{l^{ - 1}}
MB{M_B} = 111 gmol−1gmo{l^{ - 1}}
WA{W_A} = 580 g
ΔTf\Delta {T_f} = 2 K
We need to find out theWB{W_B}. Also we need to find out the van’t hoff factorii. That is:
According to the dissociation of CaCl2CaC{l_2},
CaCl2→Ca2++2Cl−CaC{l_2} \to C{a^{2 + }} + 2C{l^ - }
For this reaction, the value of van’t hoff factor ii = 3.
Substituting the all these values in the equation, we get
2=3×(1.86×WB×1000)111×5802 = 3 \times \dfrac{{\left( {1.86 \times {W_B} \times 1000} \right)}}{{111 \times 580}}
⇒WB=2×111×5801.86×3×1000\Rightarrow {W_B} = \dfrac{{2 \times 111 \times 580}}{{1.86 \times 3 \times 1000}}
⇒WB=23.07g\Rightarrow {W_B} = 23.07g
So, the amount of CaCl2CaC{l_2} is required to add so that the freezing point will be depressed by 2K is 23.07 g.

Note:
While we solve this type of problems, they will not give the value of the van't hoff factor. We need to find out this value by writing the dissociation reaction of solute and the difference between the number of moles of products and reactants will be the van’t hoff factor.