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Question: 1.355 g of a substance dissolved in 55g of \(C{H}_{2}COOH\) produced a depression in the freezing po...

1.355 g of a substance dissolved in 55g of CH2COOHC{H}_{2}COOH produced a depression in the freezing point of 0.6180C{0.618}^{0}C. Calculate the molecular weight of the substance. (Kf=3.85)({K}_{f} = 3.85).

Explanation

Solution

Hint: The freezing point of a temperature depends upon the depression in the freezing point. The depression in freezing point is the change in the freezing point of a substance. It is inversely related to the freezing point. The higher the depression, the lower will be the freezing point.

Complete step by step answer:
Freezing point can be defined as the temperature at which a liquid becomes a solid at normal atmospheric temperature. The freezing point of the solution also depends upon the depression in freezing point (δT)(\delta T). The depression in freezing point (δT)(\delta T) is defined as the decrease in the freezing point on the addition of a non-volatile solute.
The depression in freezing point in term of mass of solute and solvent as follows:
ΔTf=Kf×wB×1000MB×wB\Delta { T }_{ f }\quad =\quad { K }_{ f }\quad \times \quad \cfrac { { w }_{ B }\quad \times \quad 1000 }{ { M }_{ B }\quad \times \quad { w }_{ B } }
where, δTf\delta {T}_{f}=depression in freezing point, Kf{K}_{f}=molal freezing point depression constant of solute, wB{w}_{B}=mass of solute, MB{M}_{B}=molecular weight of solute and wA{w}_{A}=mass of solvent
We need to find the molecular weight of the solute. Therefore, the equation can be rearranged as follows.
MB=Kf×wB×1000ΔTF×wB{ M }_{ B }\quad =\quad { K }_{ f }\quad \times \quad \cfrac { { w }_{ B }\quad \times \quad 1000 }{ { \Delta T }_{ F }\quad \times \quad { w }_{ B } }
Given, δTf=0.6180C\delta {T}_{f}={0.618}^{0}C, Kf=3.85{K}_{f}=3.85, wB=1.355g{w}_{B}=1.355g , and wA=55g{w}_{A}=55g. Substituting these values in the above equation, we get
MB=3.85×1.355×10000.618×55{ M }_{ B }\quad =\quad 3.85\quad \times \quad \cfrac { 1.355\quad \times \quad 1000 }{ 0.618\quad \times \quad 55 }
    MB=153.47g/mol\implies { M }_{ B }\quad =\quad 153.47\quad g/mol
Therefore, the molecular weight of the substance is 153.47 g/mol.

Note: It is important to determine what is the solvent and what is the solute in the solution. Solute is the substance that is dissolved in a minor quantity in the solvent. And solvent is the substance which dissolves the solute.