Calculate the depression in freezing point of solution when 4\,{g} nonvolatile solute of molar mass 126\,{g}\,{mol}^{-1} dissolved in 80\,{mL} water [Cryoscopic constant of water =1.86\,{K}\,{kg}\,{mol}^{-1} ]

Correct Answer is : 0.74 K0.74\,{K}0.74K

Correct Answer is : 0.74 K0.74\,{K}0.74K

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MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Chemistry

Question : 39 of 150

Marks: +1, -0

4\,\mathrm{g}

126\,\mathrm{g}\,\mathrm{mol}^{-1}

80\,\mathrm{mL}

=1.86\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}

]

Solution:

To calculate the depression in freezing point, we use the formula derived from the concept of colligative properties:

\Delta T_f = K_f \cdot \;\frac{w_2}{M_2} \cdot \;\frac{1}{w_1}

Here:

\Delta T_f

is the depression in freezing point.

K_f

is the cryoscopic constant of the solvent (water), which is

1.86\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}

w_2

is the mass of the solute, which is

4\,\mathrm{g}

M_2

is the molar mass of the solute, which is

126\,\mathrm{g}\,\mathrm{mol}^{-1}

w_1

is the mass of the solvent, which we need to calculate from the given volume of water

80\,\mathrm{mL}

First, we convert the volume of water to mass:

Since the density of water is approximately

1\,\mathrm{g}/\mathrm{mL}

w_1 = 80\,\mathrm{mL} \times 1\,\mathrm{g}/\mathrm{mL} = 80\,\mathrm{g} = 0.08\,\mathrm{kg}

Now, substituting the values into the formula:

\Delta T_f = 1.86 \cdot \;\frac{4}{126} \cdot \;\frac{1}{0.08}

Simplify the expression step-by-step:

First calculate the fraction:

\;\frac{4}{126} = 0.03175

Then:

\Delta T_f = 1.86 \cdot 0.03175 \cdot 12.5

Next, we multiply 0.03175 by 12.5 :

0.03175 \cdot 12.5 = 0.396875

Finally:

\Delta T_f = 1.86 \cdot 0.396875 = 0.7381875 \approx 0.74\,\mathrm{K}

Therefore, the correct answer is:

Option B:

0.74\,\mathrm{K}

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