1.355 g of a substance dissolved in 55g of $C{H}_{2}COOH$ produced a depression in the freezing point of ${0.618}^{0}C$. Calculate the molecular weight of the substance. $({K}_{f} = 3.85)$.
Answer
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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 $(\delta T)$. The depression in freezing point $(\delta T)$ 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:
$\Delta { T }_{ f }\quad =\quad { K }_{ f }\quad \times \quad \cfrac { { w }_{ B }\quad \times \quad 1000 }{ { M }_{ B }\quad \times \quad { w }_{ B } }$
where, $\delta {T}_{f}$=depression in freezing point, ${K}_{f}$=molal freezing point depression constant of solute, ${w}_{B}$=mass of solute, ${M}_{B}$=molecular weight of solute and ${w}_{A}$=mass of solvent
We need to find the molecular weight of the solute. Therefore, the equation can be rearranged as follows.
${ M }_{ B }\quad =\quad { K }_{ f }\quad \times \quad \cfrac { { w }_{ B }\quad \times \quad 1000 }{ { \Delta T }_{ F }\quad \times \quad { w }_{ B } }$
Given, $\delta {T}_{f}={0.618}^{0}C$, ${K}_{f}=3.85$, ${w}_{B}=1.355g $, and ${w}_{A}=55g$. Substituting these values in the above equation, we get
${ M }_{ B }\quad =\quad 3.85\quad \times \quad \cfrac { 1.355\quad \times \quad 1000 }{ 0.618\quad \times \quad 55 }$
$\implies { M }_{ B }\quad =\quad 153.47\quad g/mol$
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.
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 $(\delta T)$. The depression in freezing point $(\delta T)$ 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:
$\Delta { T }_{ f }\quad =\quad { K }_{ f }\quad \times \quad \cfrac { { w }_{ B }\quad \times \quad 1000 }{ { M }_{ B }\quad \times \quad { w }_{ B } }$
where, $\delta {T}_{f}$=depression in freezing point, ${K}_{f}$=molal freezing point depression constant of solute, ${w}_{B}$=mass of solute, ${M}_{B}$=molecular weight of solute and ${w}_{A}$=mass of solvent
We need to find the molecular weight of the solute. Therefore, the equation can be rearranged as follows.
${ M }_{ B }\quad =\quad { K }_{ f }\quad \times \quad \cfrac { { w }_{ B }\quad \times \quad 1000 }{ { \Delta T }_{ F }\quad \times \quad { w }_{ B } }$
Given, $\delta {T}_{f}={0.618}^{0}C$, ${K}_{f}=3.85$, ${w}_{B}=1.355g $, and ${w}_{A}=55g$. Substituting these values in the above equation, we get
${ M }_{ B }\quad =\quad 3.85\quad \times \quad \cfrac { 1.355\quad \times \quad 1000 }{ 0.618\quad \times \quad 55 }$
$\implies { M }_{ B }\quad =\quad 153.47\quad g/mol$
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.
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