When 600 mL of 0.2 M HNO3 is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ________ × 10–2 °C.
(Enthalpy of neutralisation = 57 kJ mol–1 and Specific heat of water = 4.2 JK–1 g–1)
(Neglect heat capacity of flask)
(Enthalpy of neutralisation = 57 kJ mol–1 and Specific heat of water = 4.2 JK–1 g–1)
(Neglect heat capacity of flask)
Correct Answer: 54
Solution and Explanation
\(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\)
m moles 120 40 – 80 – 40
Heat liberated from reaction
= 40 × 10–3× 57 × 103 J …(1)
Heat gained by solution = \(mC\Delta T\)
Mass of solution \(m = V × d = 1000 × 1 = 1000 g\)
Heat gained by solution \(= 1000 × 4.2 × ΔT …(2)\)
From eq (1) and eq (2)
Heat liberated = Heat gained
\(40 × 10–3 × 57 × 103 = 1000 × 4.2 × ΔT\)
\(ΔT = 54 ×\) 10–2 \(°C\)
\(ΔT = 54°C\) (Rounded off to the nearest integer)
So, the answer is 54°C.
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
- The sum of all rational numbers in \( (2 + \sqrt{3})^8 \) is:
- JEE Main - 2025
- Binomial theorem
- The maximum covalency of a non-metallic group 15 element 'E' with the weakest E-E bond is:
- JEE Main - 2025
- Group 15 Elements
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
- JEE Main - 2025
- Conic sections
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
- JEE Main - 2025
- Number Systems
- For an experimental expression \( y = \frac{32.3 \times 1125}{27.4} \), where all the digits are significant. Then to report the value of \( y \), we should write:
- JEE Main - 2025
- Significant figures
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.