Molal depression constant for a solvent is $4.0\, kg\, mol^{-1}$. The depression in the freezing point of the solvent for $0.03\, mol\, kg^{-1}$ solution of $ {K2SO4}$ is : (Assume complete dissociation of the electrolyte)
- 0.12 K
- 0.36 K
- 0.18 K
- 0.24 K
The Correct Option is B
Solution and Explanation
Top Questions on Colligative Properties
- Given below are two statements:
Statement (I): NaCl is added to the ice at \( 0^\circ C \), present in the ice cream box to prevent the melting of ice cream.
Statement (II): On addition of NaCl to ice at \( 0^\circ C \), there is a depression in freezing point.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
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If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
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1.24 g of \(AX_2\) (molar mass 124 g mol\(^{-1}\)) is dissolved in 1 kg of water to form a solution with boiling point of 100.105°C, while 2.54 g of AY_2 (molar mass 250 g mol\(^{-1}\)) in 2 kg of water constitutes a solution with a boiling point of 100.026°C. \(Kb(H)_2\)\(\text(O)\) = 0.52 K kg mol\(^{-1}\). Which of the following is correct?
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- Consider the given plots of vapor pressure (VP) vs temperature (T/K). Which amongst the following options is the correct graphical representation showing \( \Delta T_f \) depression in the freezing point of solvent in a solution?
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- HA $ (aq) \rightleftharpoons H^+ (aq) + A^- (aq) $ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: $ K_f(H_2O) = 1.8 \, \text{K kg mol}^{-1} $, molality ≡ molarity
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Questions Asked in JEE Main exam
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
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Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
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- Let \( [x] \) denote the greatest integer function, and let \( m \) and \( n \) respectively be the numbers of the points, where the function \( f(x) = [x] + |x - 2| \), \( -2<x<3 \), is not continuous and not differentiable. Then \( m + n \) is equal to:
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The effect of temperature on the spontaneity of reactions are represented as: Which of the following is correct?
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Concepts Used:
Colligative Properties
Colligative Property of any substance is entirely dependent on the ratio of the number of solute particles to the total number of solvent particles but does not depend on the nature of particles. There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
Examples of Colligative Properties
We can notice the colligative properties of arrangements by going through the accompanying examples:
- On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
- In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.
Types of Colligative Properties
- Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
- Boiling point elevation: ΔTb = kb m
- Osmotic pressure: π = (n2/V) RT
- Relative lowering of vapor pressure: (Po - Ps)/Po