MHTCET 14 May 2023 Shift 1 Solved Paper

To calculate the depression in freezing point, we use the formula derived from the concept of colligative properties:
∆Tf=Kf⋅‌

w2

M2

⋅‌

1

w1

Here:
∆Tf is the depression in freezing point.
Kf is the cryoscopic constant of the solvent (water), which is 1.86K‌kg‌mol−1.
w2 is the mass of the solute, which is 4g.
M2 is the molar mass of the solute, which is 126gmol−1.
w1 is the mass of the solvent, which we need to calculate from the given volume of water 80‌mL.
First, we convert the volume of water to mass:
Since the density of water is approximately 1g∕mL,
w1=80‌mL×1g∕mL=80g=0.08‌kg
Now, substituting the values into the formula:
∆Tf=1.86⋅‌

4

126

⋅‌

1

0.08

Simplify the expression step-by-step:
First calculate the fraction:
‌

4

126

=0.03175
Then:
∆Tf=1.86⋅0.03175⋅12.5
Next, we multiply 0.03175 by 12.5 :
0.03175⋅12.5=0.396875
Finally:
∆Tf=1.86⋅0.396875=0.7381875≈0.74K
Therefore, the correct answer is:
Option B: 0.74K