Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at $-24^\circ C$ is ____ kg (Molar mass in g mol$^{-1}$ for ethylene glycol = 62, $K_f$ of water = 1.86 K kg mol$^{-1}$).
Correct Answer: 15
Solution and Explanation
Step-by-step Calculation
The depression in freezing point (\( \Delta T_f \)) is given by:
\[\Delta T_f = K_f \times m\]where:
\( \Delta T_f = 24^\circ\text{C} \) (since the freezing point is to be lowered to \( -24^\circ\text{C} \))
\( K_f = 1.86 \, \text{K kg mol}^{-1} \) (cryoscopic constant of water)
\( m \) is the molality of the solution.
Rearranging the formula to find molality:
\[m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86} \approx 12.903 \, \text{mol kg}^{-1}\]
Calculating the Mass of Ethylene Glycol:
Molality (\( m \)) is defined as:
\[m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\]
Let \( n \) be the number of moles of ethylene glycol. Therefore:
\[n = m \times \text{mass of solvent} = 12.903 \times 18.6 \approx 240.9958 \, \text{mol}\]
The mass of ethylene glycol is given by:
\[\text{Mass of ethylene glycol} = n \times \text{molar mass of ethylene glycol}\]
Substituting the known values:
\[\text{Mass of ethylene glycol} = 240.9958 \times 62 \approx 14,941.74 \, \text{g} \approx 15 \, \text{kg}\]
Conclusion:The mass of ethylene glycol required is approximately \( 15 \, \text{kg} \).
Top Questions on Solutions
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this concept is valid in ecological systems (isotopic mixtures of an element, hydrocarbons mixtures, etc.). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This law is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
Answer the following questions:
(a) Give one example of miscible liquid pair which shows negative deviation from Raoult’s law. What is the reason for such deviation?
(b) (i) State Raoult’s law for a solution containing volatile components.
OR
(ii) Raoult’s law is a special case of Henry’s law. Comment.
(c) Write two characteristics of an ideal solution.- A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water \(K_f\) and \(K_b\) are 1.86 K kg mol\(^{-1}\) and 0.512 K kg mol\(^{-1}\) respectively.
- Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example.
- Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? - An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because __________
Questions Asked in JEE Main exam
- Consider a long thin conducting wire carrying a uniform current \( I \). A particle having mass \( M \) and charge \( q \) is released at a distance \( a \) from the wire with a speed \( v_0 \) along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance \( x \) from the wire. The value of \( x \) is:
- JEE Main - 2025
- Magnetic Field
- Two equal sides of an isosceles triangle are along \( -x + 2y = 4 \) and \( x + y = 4 \). If \( m \) is the slope of its third side, then the sum of all possible distinct values of \( m \) is:
- JEE Main - 2025
- 3D Geometry
- Let the position vectors of the vertices A, B, and C of a tetrahedron ABCD be \( \hat{i} + 2\hat{j} + \hat{k} \), \( \hat{i} + 3\hat{j} - 2\hat{k} \), and \( 2\hat{i} + \hat{j} - \hat{k} \) respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is \( \frac{\sqrt{10}}{3} \) and the volume of the tetrahedron is \( \frac{\sqrt{805}}{6\sqrt{2}} \), then the position vector of E is:
- JEE Main - 2025
- Geometry and Vectors
- If \( x_1, x_2, x_3, x_4 \) are in GP (Geometric Progression), then we subtract 2, 4, 7, and 8 from \( x_1, x_2, x_3, x_4 \) respectively, then the resultant numbers are in AP (Arithmetic Progression). Then the value of \( \frac{1}{24} (x_1 \cdot x_2 \cdot x_3 \cdot x_4) \) is:
- JEE Main - 2025
- Arithmetic Progression
- Let the area of a triangle \( \triangle PQR \) with vertices \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \) be 35 square units. If its orthocenter and centroid are \( O(2, \frac{14}{5}) \) and \( C(c, d) \) respectively, then \( c + 2d \) is equal to:
- JEE Main - 2025
- Geometry and Vectors