
1.00 g of non-electrolyte solute dissolved in 50 g of benzene lowered the point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 Kkg/mol. Find the molar mass of the solute.
Answer
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Hint: The answer to this question is based on the formula which is used to calculate the depression in the freezing point which is denoted by $\Delta {{T}_{f}} = \dfrac{1000\times w\times {{k}_{f}}}{M\times W}$ where ${{k}_{f}}$ is the depression in freezing point constant.
Complete step by step answer:
We are familiar with the colligative properties of the compounds from the physical chemistry concept and also terms and definitions relating to it.
Now, let us see the meaning of the depression in freezing point and how it is calculated.
- Depression in freezing point is the decrease in the freezing point of the solvent when a non-volatile solute is added to it or in simple words it can be said that the freezing point of a solution is less than that of freezing point of the pure solvent. Temperature plays an important role here.
- Now, let us calculate the above given question according to which the data given as,
Mass of the solute (non-electrolyte) = $w$ = 1 g
Mass of the solvent that is $W$ = 50 g
Depression in the freezing point is given and is $\Delta {{T}_{f}}$ = 5.12 K kg/mol
Also, freezing point constant is given as ${{k}_{f}}$ = 0.40 K
Now, the depression in the freezing point is given by the formula $\Delta {{T}_{f}} = \dfrac{1000\times w\times {{k}_{f}}}{M\times W}$
Rearranging this equation we get, $M=\dfrac{1000\times w\times {{k}_{f}}}{\Delta {{T}_{f}}\times W}$
Substituting the values, we can get the value of the molar mass of solute.
Therefore, $M = \dfrac{1000\times 1\times 5.12}{0.40\times 50}$
\[\Rightarrow M = 256g/mol\]
Therefore, the correct answer is 256 g/mol which is the molar mass of the solute.
Note: Note that depression in the freezing point is also known as cryoscopy which measures the lowered freezing points in liquid by dissolved substances for determination of molecular weights of solutes and several properties of solution and the instrument used for this is called cryoscope.
Complete step by step answer:
We are familiar with the colligative properties of the compounds from the physical chemistry concept and also terms and definitions relating to it.
Now, let us see the meaning of the depression in freezing point and how it is calculated.
- Depression in freezing point is the decrease in the freezing point of the solvent when a non-volatile solute is added to it or in simple words it can be said that the freezing point of a solution is less than that of freezing point of the pure solvent. Temperature plays an important role here.
- Now, let us calculate the above given question according to which the data given as,
Mass of the solute (non-electrolyte) = $w$ = 1 g
Mass of the solvent that is $W$ = 50 g
Depression in the freezing point is given and is $\Delta {{T}_{f}}$ = 5.12 K kg/mol
Also, freezing point constant is given as ${{k}_{f}}$ = 0.40 K
Now, the depression in the freezing point is given by the formula $\Delta {{T}_{f}} = \dfrac{1000\times w\times {{k}_{f}}}{M\times W}$
Rearranging this equation we get, $M=\dfrac{1000\times w\times {{k}_{f}}}{\Delta {{T}_{f}}\times W}$
Substituting the values, we can get the value of the molar mass of solute.
Therefore, $M = \dfrac{1000\times 1\times 5.12}{0.40\times 50}$
\[\Rightarrow M = 256g/mol\]
Therefore, the correct answer is 256 g/mol which is the molar mass of the solute.
Note: Note that depression in the freezing point is also known as cryoscopy which measures the lowered freezing points in liquid by dissolved substances for determination of molecular weights of solutes and several properties of solution and the instrument used for this is called cryoscope.
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