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Sedbober [7]
3 years ago
13

A new compound was recently discovered and found to have an atomic weight of 342.38 amu. This element has two isotopes, the ligh

ter of which has a mass of 340.91 amu and an abundance of 68.322%. What is the mass of the heavier isotope?
Chemistry
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer:

The mass of heavier isotope is 345.6 amu.

Explanation:

Given data:

atomic wight of compound = 342.38 amu

lighter isotope mass = 340.91 amu

abundance of lighter isotope = 68.322%

mass of heavier isotope = ?

Solution:

average atomic mass = ( % age abundance of lighter isotope × its atomic mass) + (% age abundance of heavier isotope × its atomic mass)  / 100

percentage of heavier isotope = 100- 68.322 = 31.678

Now we will put the values in formula.

342.38  = (68.322× 340.91) + (31.678 × X) / 100

342.38 = 23291.65302 +  (31.678 × X) / 100

342.38 × 100 = 23291.65302 +  (31.678 × X)

34238 -23291.65302 =  (31.678 × X)

10946.35 / 31.678 = X

345.6 = X

The mass of heavier isotope is 345.6 amu.

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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
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Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

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As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

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Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

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Substitute the values into above formula as follows.

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Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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