\(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\)
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.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the frequency of the supply across resistor \( R \) and inductor \( L \). If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In light of the above statements, choose the most appropriate answer from the options given below:
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.
The solutions can be classified into three types:
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types: