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Chapter 1: Problem 61

During the conversion of \(\mathrm{NH}_{2} \mathrm{OH} \longrightarrow \mathrm{N}_{2} \mathrm{O}\), the equivalent weight of \(\mathrm{NH}_{2} \mathrm{OH}\) is (mol. wt. of \(\mathrm{NH}_{2} \mathrm{OH}\) is \(\mathrm{M}\) ) (1) \(M\) (2) \(\mathrm{M} / 2\) (3) \(\mathrm{M} / 4\) (4) M/5 (5) \(2 \mathrm{M}\)

Short Answer

Expert verified
\( \text{M} / 2 \)

Step by step solution

01

- Understanding Equivalent Weight

Equivalent weight is defined as the molecular weight divided by the number of moles of electrons lost or gained in the conversion. Here, we need to determine the change in oxidation numbers to find the number of electrons transferred.
02

- Oxidation Numbers of \(\text{NH}_{2} \text{OH}\)

Calculate the oxidation number of nitrogen in \(\text{NH}_{2} \text{OH}\). Hydrogen has an oxidation state of +1, and oxygen has -2. The sum of the oxidation states in \(\text{NH}_{2} \text{OH}\) must equal zero: \(\text{N} + 2(\text{H(+1)}) + \text{O(-2)} + \text{H(+1)} = 0\). Solving for \(\text{N}\), \(\text{N} = -1\).
03

- Oxidation Numbers of \(\text{N}_{2} \text{O}\)

Calculate the oxidation number of nitrogen in \(\text{N}_{2} \text{O}\). Oxygen has an oxidation state of -2, and the sum of oxidation states must equal zero for a neutral molecule. This results in 2n - 2 = 0, making n = +1. But since nitrogen exists in pairs in \(\text{N}_{2} \text{O}\), the average oxidation state is +1.
04

- Change in Oxidation States

Determine the change in oxidation states for nitrogen. Initially, nitrogen is at -1, and it changes to +1: \(-1 \to +1\). The change is an increase of 2 electrons (each nitrogen changes from -1 to +1, totaling 2).
05

- Equivalent Weight Calculation

Using the definition, equivalent weight = molecular weight / number of moles of electrons transferred. The molecular weight of \(\text{NH}_{2} \text{OH}\) is \(\text{M}\), and the number of moles of electrons transferred is 2. Therefore, the equivalent weight = \(\text{M} / 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Numbers
Oxidation numbers are a way to keep track of electrons in chemical reactions. They help us understand how electrons are gained or lost during a reaction. Each element in a molecule is assigned an oxidation number based on certain rules. For example, in \(\text{NH}_{2} \text{OH}\), hydrogen generally has an oxidation number of +1, while oxygen has an oxidation number of -2. The sum of the oxidation numbers in a neutral molecule must always be zero. Therefore, for \(\text{NH}_{2} \text{OH}\), we get: \(\text{N} + 2(\text{H}(+1)) + \text{O}(-2) + \text{H}(+1) = 0\). Solving this equation, we find that the oxidation number of nitrogen (N) is -1.
Chemical Reactions
Chemical reactions involve the rearrangement of atoms and the transfer of electrons between substances. To understand the conversion of \(\text{NH}_{2} \text{OH} \to \text{N}_{2} \text{O}\), we need to look at the changes in oxidation numbers. In \(\text{N}_{2} \text{O}\), the oxidation number of nitrogen is calculated considering that oxygen’s oxidation number is -2, and the sum of oxidation states in a neutral molecule should be zero. So we have: \(2n - 2 = 0\), which simplifies to \(n = +1\). Thus, each nitrogen in \(\text{N}_{2} \text{O}\) has an average oxidation number of +1. Understanding these changes provides insight into the number of electrons transferred during the reaction.
Molecular Weight
The molecular weight (also known as molar mass) is the sum of the atomic weights of all the atoms in a molecule. For \(\text{NH}_{2} \text{OH}\), suppose the molecular weight is denoted as \(\text{M}\). When calculating the equivalent weight, we use the molecular weight and the number of moles of electrons transferred. In the given reaction, nitrogen goes from an oxidation state of -1 to +1, implying a transfer of 2 electrons. So, the equivalent weight is the molecular weight divided by the number of moles of electrons transferred. Thus, the equivalent weight of \(\text{NH}_{2} \text{OH}\) in this reaction is \(\frac{\text{M}}{2}\).

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Most popular questions from this chapter

A circular disc of radius \(\mathrm{R}\) rolls without slipping on a rough horizontal surface. At the instant shown its linear velocity is \(\mathrm{V}\), linear acceleration a, angular velocity \(\omega\) and angular acceleration \(\alpha\). Four points \(\mathrm{A}\), B, C and D lie on its circumference such that the diameter AC is vertical \& BD horizontal then choose the correct options. (1) \(\mathrm{V}_{\mathrm{B}}=\sqrt{\mathrm{V}^{2}+(\mathrm{R} \omega)^{2}}\) (2) \(V_{C}=V+R \omega\) (3) \(\mathrm{a}_{\mathrm{A}}=\sqrt{(\mathrm{a}-\mathrm{R} \alpha)^{2}+\left(\omega^{2} \mathrm{R}\right)^{2}}\) (4) \(\mathrm{a}_{\mathrm{D}}=\sqrt{\left(\mathrm{a}+\omega^{2} \mathrm{R}\right)^{2}+(\mathrm{R} \alpha)^{2}}\)

A steel wire is rigidly fixed along diameter of aluminium ring of radius \(\mathrm{R}\) as shown. Linear expansion coefficient of steel is half of linear expansion coefficient for aluminium. If temperature of system is increased by \(\Delta \theta\) then the thermal stress developed in steel wire is: \(\left(\alpha_{\mathrm{Al}}\right.\) is linear expansion coefficient for aluminium and Young's modulus for steel is \(Y\) ) (1) more than \(\mathrm{R} \alpha_{\mathrm{Al}} \Delta \theta \mathrm{Y}\) (2) less then R \(\alpha_{\mathrm{Al}} \Delta \theta \mathrm{Y}\) (3) Equal to \(R \alpha_{\mathrm{Al}} \Delta \theta Y\) (4) equal to \(2 R \alpha_{A 1} \Delta \theta Y\) (5) equal to \(4 \mathrm{R} \alpha_{\mathrm{Al}} \Delta \theta \mathrm{Y}\)

Ultraviolet light of wavelength \(\lambda_{1}\) and \(\lambda_{2}\left(\lambda_{2}>\lambda_{1}\right)\) when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies \(E_{1}\) and \(E_{2}\) respectively. The value of planck's constant can be found from the relation. (1) \(\mathrm{h}=\frac{1}{\mathrm{c}}\left(\lambda_{2}-\lambda_{1}\right)\left(\mathrm{E}_{1}-\mathrm{E}_{2}\right)\) (2) \(\mathrm{h}=\frac{1}{\mathrm{c}}\left(\lambda_{1}+\lambda_{2}\right)\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right)\) (3) \(\mathrm{h}=\frac{\left(\mathrm{E}_{1}-\mathrm{E}_{2}\right) \lambda_{1} \lambda_{2}}{\mathrm{c}\left(\lambda_{2}-\lambda_{1}\right)}\) (4) \(\mathrm{h}=\frac{\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right) \lambda_{1} \lambda_{2}}{\mathrm{c}\left(\lambda_{1}+\lambda_{2}\right)}\) (5) \(\mathrm{h}=\frac{\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right) \lambda_{1} \lambda_{2}}{3 \mathrm{c}\left(\lambda_{2}-\lambda_{1}\right)}\)

The length of latus rectum of the ellipse whose focus is \(\mathrm{S}(-1,1)\), the corresponding directrix is \(x-y+3=0\) and the eccentricity is \(\frac{1}{2}\), is equal to (1) \(\frac{1}{2}\) (2) \(\frac{1}{2 \sqrt{2}}\) (3) \(\frac{1}{\sqrt{2}}\) (4) \(\frac{1}{4}\)

The vapour pressure of the solution of two liquids \(\mathrm{A}\left(\mathrm{p}^{\circ}=80 \mathrm{~mm}\right)\) and \(\mathrm{B}\left(\mathrm{p}^{\circ}=120 \mathrm{~mm}\right)\) is found to be \(100 \mathrm{~mm}\) when \(\mathrm{x}_{\mathrm{A}}=0.4\). The result shows that : (1) solution exhibits ideal behaviour (2) \(\Delta \mathrm{H}_{\text {solution }}<0\) (3) solution shows negative deviations (4) solution will show positive deviations for lower concentration and negative deviations for higher concentrations.
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