back to directory
NEET CHEMISTRYSolutionsMedium

Question

1.00 g1.00 \text{ g} of a non-electrolyte solute (molar mass 250 g/mol250 \text{ g/mol}) was dissolved in 51.2 g51.2 \text{ g} of benzene. If the freezing point depression constant, KfK_f of benzene is 5.12 K kg mol15.12 \text{ K kg mol}^{-1}, the freezing point of benzene will be lowered by:

A

0.4 K0.4 \text{ K}

B

0.3 K0.3 \text{ K}

C

0.5 K0.5 \text{ K}

D

0.2 K0.2 \text{ K}

Step-by-Step Solution

Given: Mass of solute (w2w_2) = 1.00 g1.00 \text{ g} Molar mass of solute (M2M_2) = 250 g/mol250 \text{ g/mol} Mass of solvent (benzene, w1w_1) = 51.2 g=0.0512 kg51.2 \text{ g} = 0.0512 \text{ kg} Freezing point depression constant (KfK_f) = 5.12 K kg mol15.12 \text{ K kg mol}^{-1}

The depression in freezing point (ΔTf\Delta T_f) is given by the formula: ΔTf=Kf×m\Delta T_f = K_f \times m Where mm is the molality of the solution. m=w2M2×w1 (in kg)=1.00250×0.0512=112.8 mm = \frac{w_2}{M_2 \times w_1 \text{ (in kg)}} = \frac{1.00}{250 \times 0.0512} = \frac{1}{12.8} \text{ m}

Now, substitute the values into the equation: ΔTf=5.12×112.8=0.4 K\Delta T_f = 5.12 \times \frac{1}{12.8} = 0.4 \text{ K}

Therefore, the freezing point of benzene will be lowered by 0.4 K0.4 \text{ K}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSolutionsnonelectrolytesolutedissolvedbenzenefreezing

More Solutions Questions

View all

Isotonic solutions have the same:

A.Vapour pressure
B.Freezing temperature
C.Osmotic pressure
D.Boiling temperature
EasySolve

A solution containing $10 \text{ g/dm}^3$ of urea (molecular mass $= 60 \text{ g mol}^{-1}$) is isotonic with a $5 \%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is:

A.25 g mol⁻¹
B.300 g mol⁻¹
C.350 g mol⁻¹
D.200 g mol⁻¹
EasySolve

Which colligative property provides the most accurate method for determining the molecular weight of proteins and polymers?

A.Osmotic pressure
B.Lowering of vapour pressure
C.Lowering of freezing point
D.Elevation in boiling point
EasySolve

The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in solution is 0.2. What would be the mole fraction of the solvent if the decrease in vapour pressure is 20 mm of Hg?

A.0.2
B.0.4
C.0.6
D.0.8
MediumSolve

Which one is not equal to zero for an ideal solution?

A.$\Delta H_{\text{mix}}$
B.$\Delta S_{\text{mix}}$
C.$\Delta V_{\text{mix}}$
D.$\Delta P = P_{\text{Observed}} - P_{\text{Raoult}}$
EasySolve

The freezing point depression constant ($K_f$) of benzene is $5.12 \text{ K kg mol}^{-1}$. The freezing point depression for the solution of molality $0.078 \text{ m}$ containing a non-electrolyte solute in benzene is:

A.0.80 K
B.0.40 K
C.0.60 K
D.0.20 K
EasySolve

The molarity of a $0.2 \text{ N } Na_2CO_3$ solution will be:

A.0.05 M
B.0.2 M
C.0.1 M
D.0.4 M
EasySolve

Of the following $0.10 \text{ m}$ aqueous solutions, which one will exhibit the largest freezing point depression?

A.$\text{KCl}$
B.$\text{C}_6\text{H}_{12}\text{O}_6$
C.$\text{Al}_2(\text{SO}_4)_3$
D.$\text{K}_2\text{SO}_4$
EasySolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →