Tag: determination of atomic and isotopic mass

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Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

There are two isotopes of an element with atomic mass $z$. Heavier one has atomic mass $z+2$ and lighter one has $z-1$, then an abundance of lighter one is:

  1. $66.6\%$
  2. $96.7\%$
  3. $6.67\%$
  4. $33.3\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Abundance of an isotope can be calculated as :


Let abundance of isotope with atomic mass $Z+2$ is x
Then abundance of isotope $Z-1$ is $1-x$

$(Z+2)x+(Z-1)(1-x)\quad =Z$

On solving this, we get
$x=\dfrac { 1 }{ 3 } $
Therefore,% abundance of isotope with atomic no.$Z+2$ is =$=\dfrac { 1 }{ 3 } \times 100=33.33$%
Therefore,% abundance isotope with atomic no. $Z-1$ is =$100-33.33=67.66%$%
Therefore isotope with atomic $Z-1$ is higher in % abundance

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

$ _{ 17 }^{ 35 }{ Cl }$ and $ _{ 17 }^{ 37 }{ Cl }$ are two isotopes of chlorine. If average atomic mass is $35.5$ then ratio of these two isotopes is:

  1. $35 : 37$
  2. $1 : 3$
  3. $3 : 1$
  4. $2 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solution:- (C) $3 : 1$

Let $x$ and $y$ be the fraction of ${ _{17}^{35}{Cl}}$ and ${ _{17}^{37}{Cl}}$ in ${ _{17}^{35.5}{Cl}}$.
Therefore,
Average at. mass $= \cfrac{x \times 35 + y \times 37}{x + y}$
$35.5 = \cfrac{35x + 37y}{x+y}$
$\Rightarrow 35.5 x + 35.5 y = 35x + 37y$
$\Rightarrow 35.5 x - 35x = 37 y - 35.5 y$
$\Rightarrow 0.5 x = 1.5 y$
$\Rightarrow \cfrac{x}{y} = \cfrac{1.5}{0.5} = \cfrac{3}{1}$
Hence the ratio of ${ _{17}^{35}{Cl}}$ and ${ _{17}^{37}{Cl}}$ in ${ _{17}^{35.5}{Cl}}$ is $3 : 1$.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

If isotopic distribution of $ C-12 $ and $ C-14 $ is  98 %  and  2 %  respectively, what would be the number of $ C-14 $ isotope in $ 12 gm $  carbon sample?

  1. $ 1.032 \times 10^{22} $
  2. $ 3.01 \times 10^{23} $
  3. $ 5.88 \times 10^{23} $
  4. $ 6.02 \times 10^{23} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1 mole of carbon contains $N _A$ molecules.


$14g$ of $C-14$ contains $6.023 \times 10^{23}$ atoms

$1g$ of $C-14= \dfrac{6.023 \times 10^{23}}{14}$


$12g$ of $C-14$ atom contains $=\cfrac {6.023 \times 106{23}\times 12}{14}$

                                                  $=5.16 \times 10^{23}$ atoms

Now $2$% of $5.16 \times 10^{23}=\cfrac {5.16 \times 10^{23}\times 2}{100}$

                                         $=1.032 \times 10^{22}$ atoms .

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

Chlorine has two naturally occurring isotopes, $^{35}Cl$ and $^{37}Cl$. If the atomic mass of Cl is 35.5 the ratio of natural abundance of $^{35}Cl$ and $^{37}Cl$ is closest to :

  1. $ 3 : 5 $
  2. $3 : 1$
  3. $2 : 5 $
  4. $4 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let x be the fraction of Cl-35. Then 35x + 37(1-x) = 35.5. 35x + 37 - 37x = 35.5. -2x = -1.5. x = 0.75. The ratio of Cl-35 to Cl-37 is 0.75 / 0.25 = 3:1.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

A sample of perfect gas that initially occupies $15.0L$ at $300K$ and $1.0$ bar is compressed isothermally. To what volume must the gas be compressed to reduce its entropy by $5.0J/K$? $\left[ \ln { 0.36 } =-1.0,\ln { 2.7 } =1.0 \right] $

  1. $5.4L$
  2. $8.22L$
  3. $40.5L$
  4. $5.56L$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an isothermal process, delta S = nR ln(V2/V1). Since nR = PV/T = (1 bar * 15 L) / 300 K = 0.05 L*bar/K. Converting to SI units (1 bar = 10^5 Pa, 1 L = 10^-3 m^3), nR = 5 J/K. Then -5 = 5 ln(V2/15). ln(V2/15) = -1. V2/15 = e^-1 = 0.36. V2 = 15 * 0.36 = 5.4 L.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

A system undergoes a process in which the entropy change is $+5.51J{K}^{-1}$. During the process, $1.50kJ$ of heat is added to the system at $300K$. The correct information regarding the process is

  1. the process is thermodynamically reversible

  2. the process is thermodynamically irreversible

  3. the process may or may not be thermodynamically reversible

  4. the process must be isobaric
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a reversible process, delta S = q_rev / T. Here, q/T = 1500 J / 300 K = 5 J/K. Since the actual entropy change (5.51 J/K) is greater than q/T (5 J/K), the process is irreversible.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

The table shows the numbers of particles present in the nuclei of four atoms or ions.

protons neutrons electronic structure
$1$ $18$ $22$ $2, 8, 8$
$2$ $19$ $20$ $2, 8, 8$
$3$ $19$ $21$ $2, 8, 8, 1$
$4$ $20$ $20$ $2, 8, 8, 2$

Which two particles belong to the same element?

  1. $1$ and $2$
  2. $1$ and $4$
  3. $2$ and $3$
  4. $2$ and $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\text{An element always consist same number of proton and electrons.}$

$\text{Number of neutrons can be changed in an element based on its isotrops.}$
$\text{Option C is correct.}$

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

The average atomic mass of copper is $63.546$ amu .Natural copper consists of two iostopes: $^{63} Cu$ and $^{65} Cu$.Their natural abundances are $69.09\%$ and $30.91\%$ respectively. If the mass of $^{63} Cu$ isotope is $62.9298$ amu ,What is the mass of $^{65} Cu$ isotope?

  1. 64.9000

  2. 65.1233

  3. 64.9233

  4. 65.1933
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Average mass = (mass1 * abundance1) + (mass2 * abundance2). 63.546 = (62.9298 * 0.6909) + (mass2 * 0.3091). 63.546 = 43.4779 + 0.3091 * mass2. 20.0681 = 0.3091 * mass2. mass2 = 64.9243.