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Chapter 1: Problem 2

The average molar mass of chlorine is \(35.5 \mathrm{~g} \mathrm{~mol}^{-1}\). The ratio of \({ }^{35} \mathrm{Cl}\) to \({ }^{37} \mathrm{Cl}\) in naturally occrring chlorine is close to: (a) \(4: 1\) (b) \(3: 1\) (c) \(2: 1\) (d) \(1: 1\)

Short Answer

Expert verified
The ratio is approximately 3:1, option (b).

Step by step solution

01

Understand the Given

The problem states that the average molar mass of chlorine is 35.5 g/mol. Chlorine exists as two isotopes: \( ^{35}\text{Cl} \) with a mass of 35 g/mol and \( ^{37}\text{Cl} \) with a mass of 37 g/mol. We need to find the ratio of these two isotopes that results in the average molar mass.
02

Set Up Equation for Average Molar Mass

Let \( x \) be the fraction of \( ^{35}\text{Cl} \) and \( (1-x) \) be the fraction of \( ^{37}\text{Cl} \). The equation for the average molar mass is:\[ 35x + 37(1-x) = 35.5 \]
03

Solve the Equation

Expand and simplify the equation: \[ 35x + 37 - 37x = 35.5 \]Combine like terms:\[ -2x + 37 = 35.5 \]Subtract 37 from both sides:\[ -2x = -1.5 \]Divide both sides by -2 to solve for \( x \):\[ x = 0.75 \]
04

Calculate the Ratio

Since \( x = 0.75 \), the fraction of \( ^{35}\text{Cl} \) is 0.75, and \( 1-x = 0.25 \) for \( ^{37}\text{Cl} \). Therefore, the ratio of \( ^{35}\text{Cl} \) to \( ^{37}\text{Cl} \) is \( 0.75 : 0.25 \).Simplify the ratio by dividing both terms by 0.25:\[ 0.75/0.25 : 0.25/0.25 = 3:1 \]
05

Compare with Options

Compare the calculated ratio \( 3:1 \) with the given options. The closest match is option \( (b) \) \( 3:1 \).

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Key Concepts

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

Average Molar Mass
The average molar mass is a weighted mean that takes into account the abundance of different isotopes of an element.Each isotope of an element has a slightly different mass due to varying numbers of neutrons in the nucleus.
For chlorine, the average molar mass is given as 35.5 g/mol. This figure represents the combined mass of its isotopes in natural abundance.
Calculating the average molar mass involves summing the products of the proportion of each isotope and its respective mass. Here, the calculation considers the masses of the isotopes
  • \(^{35} ext{Cl}\), which has a molar mass of 35 g/mol, and
  • \(^{37} ext{Cl}\), which has a molar mass of 37 g/mol.
The ratio of these isotopes, based on natural abundance, is determined by the average molar mass.
Chlorine Isotopes
Chlorine naturally occurs as a mixture of two stable isotopes:
  • \(^{35} ext{Cl}\)
  • \(^{37} ext{Cl}\)
These isotopes differ in the number of neutrons, which results in different atomic masses. However, they share the same number of protons, qualifying them as isotopes of the same element.
The natural abundances of these isotopes in chlorine determine the calculated average molar mass. \(^{35} ext{Cl}\) is significantly more abundant, making up approximately 75% of naturally occurring chlorine.Conversely, \(^{37} ext{Cl}\) constitutes around 25%.
This higher amount of \(^{35} ext{Cl}\) explains why the average molar mass of chlorine is closer to 35 than to 37.
Molar Mass Calculation
In molar mass calculations involving isotopes, it's essential to account for the abundance and mass of each isotope. The general equation used is\[m_{avg} = m_1x_1 + m_2x_2 + \cdots + m_nx_n \]where
  • \(m_{avg}\) is the average molar mass,
  • \(m_n\) is the molar mass of each isotope, and
  • \(x_n\) is the relative fraction (abundance) of each isotope.
For chlorine, with \(^{35} ext{Cl}\) making up 75% and \(^{37} ext{Cl}\) making up 25%:\[35(0.75) + 37(0.25) = 35.5\]This equation means that we multiply the percentage abundance by the atomic mass of each isotope and sum the results.
This approach allows us to precisely determine the average molar mass by reflecting the actual natural occurrence of isotopes in an element.

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Most popular questions from this chapter

A mixture contains \(\mathrm{NaCl}\) and unknown chloride MCl. (i) \(1 \mathrm{~g}\) of this is dissolved in water. Excess of acidified \(\mathrm{AgNO}_{3}\) solution is added to it. \(2.567 \mathrm{~g}\) of white ppt. is formed. (ii) \(1 \mathrm{~g}\) of original mixture is heated to \(300^{\circ} \mathrm{C}\). Some vapours come out which are absorbed in acidified \(\mathrm{AgNO}_{3}\) solution, \(1.341 \mathrm{~g}\) of white precipitate was obtained. Find the molecular weight of unknown chloride. [1980]

Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve \(10 \mathrm{ppm}\) of iron in \(100 \mathrm{~kg}\) of wheat is .Atomic weight: \(\mathrm{Fe}=55.85 ; \mathrm{S}=32.00 ; \mathrm{O}=\) \(16.00\) [Main Jan. 08, 2020 (I)]

(a) One litre of a sample of hard water contains \(1 \mathrm{mg}\) of \(\mathrm{CaCl}_{2}\) and 1 \(\mathrm{mg}\) of \(\mathrm{MgCl}_{2}\). Find the total hardness in terms of parts of \(\mathrm{CaCO}_{3}\) per \(10^{6}\) parts of water by weight. (b) A sample of hard water contains \(20 \mathrm{mg}\) of \(\mathrm{Ca}^{++}\)ions per litre. How many milli-equivalent of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) would be required to soften 1 litre of the sample? (c) \(1 \mathrm{~g}\) of \(\mathrm{Mg}\) is burnt in a closed vessel which contains \(0.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants? (iii) How may milliliters of \(0.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) will dissolve the residue in the vessel.

A sugar syrup of weight \(214.2 \mathrm{~g}\) contains \(34.2 \mathrm{~g}\) of sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\). Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup.

A solution of two components containing \(n_{1}\) moles of the \(1^{\text {st }}\) component and \(n_{2}\) moles of the \(2^{\text {nd }}\) component is prepared. \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\) are the molecular weights of component 1 and 2 respectively. If \(d\) is the density of the solution in \(g \mathrm{~mL}^{-1}, \mathrm{C}_{2}\) is the molarity and \(x_{2}\) is the mole fraction of the \(2^{\text {nd }}\) component, then \(\mathrm{C}_{2}\) can be expressed as: (a) \(\mathrm{C}_{2}=\frac{1000 x_{2}}{\mathrm{M}_{1}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (b) \(\mathrm{C}_{2}=\frac{d x_{2}}{\mathrm{M}_{2}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (c) \(\mathrm{C}_{2}=\frac{1000 d x_{2}}{\mathrm{M}_{1}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (d) \(\mathrm{C}_{2}=\frac{d x_{1}}{\mathrm{M}_{2}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\)
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