Question:
How many grams of methyl alcohol should be added to $10$ litre tank of water to prevent its freezing at $268\, K$ ? $\left(K_{f} \, for\, water\, is \,1.86 K\, kg \, mol^{-1} \right)$
How many grams of methyl alcohol should be added to $10$ litre tank of water to prevent its freezing at $268\, K$ ? $\left(K_{f} \, for\, water\, is \,1.86 K\, kg \, mol^{-1} \right)$
Updated On: Aug 19, 2024
- 880.07 g
- 899.04 g
- 886.02 g
- 868.06 g
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
$\Delta T_{f} =K_{f} m$
where m = molality
273 - 268= 1.86 x $\frac{w}{M\times V}$
$5=1.86 \times\frac{w}{32\times10}$
$W=\frac{5\times32\times10}{1.86}$
$=860.2 \approx 868.06 g$
where m = molality
273 - 268= 1.86 x $\frac{w}{M\times V}$
$5=1.86 \times\frac{w}{32\times10}$
$W=\frac{5\times32\times10}{1.86}$
$=860.2 \approx 868.06 g$
Was this answer helpful?
0
0
Top Questions on Solutions
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- 0.1 M solution of KI reacts with excess of \( \text{H}_2\text{SO}_4 \) and KIO\(_3\), according to the equation: \[ 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O \]Identify the correct statements:
- (A) 200 mL KI solution reacts with 0.004 mol of KIO3
- (B) 200 mL KI solution reacts with 0.006 mol of H2SO4
- (3) 0.5 L KI solution produced 0.005 mol of I2
- (4) Equivalent weight of KIO3 is equal to Molecular weight / 5
- Which of the following binary mixture does not show the behavior of minimum boiling azeotropes?
- 10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is :
- 2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given: Ebullioscopic constant of water = $ 0.52 \, \text{K kg mol}^{-1} $)
View More Questions
Questions Asked in JEE Main exam
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____.
- JEE Main - 2025
- Magnetic Field
- A physical quantity \( Q \) is related to four observables \( a \), \( b \), \( c \), and \( d \) as follows: \[ Q = \frac{a b^4}{c d^2} \] Where:
- \( a = (60 \pm 3) \, \text{Pa} \),
- \( b = (20 \pm 0.1) \, \text{m} \),
- \( c = (40 \pm 0.2) \, \text{N·s/m}^2 \),
- \( d = (50 \pm 0.1) \, \text{m} \).
Then the percentage error in \( Q \) is:- JEE Main - 2025
- Error analysis
- 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:
- JEE Main - 2025
- osmosis
View More Questions
Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.