
The number of potassium atoms present in 117 g of potassium sample (Molecular Weight = 39 g/mole) are:
A. \[39 \times 6.023 \times {10^{23}}\]atoms
B. \[117 \times 6.023 \times{10^{23}}\]atoms
C. \[4 \times 6.023 \times {10^{23}}\]atoms
D. \[3 \times 6.023 \times {10^{23}}\]atoms
Answer
125.7k+ views
Hint: The molecular weight of potassium is 39 g/mole . This means the weight of 1 mole potassium which contains \[6.023{\text{ }} \times {\text{ }}{10^{23}}\] number of atoms is 39 g. To calculate the number of potassium in 117 grams of potassium, you should know the mole concept.
Formula used: \[{\text{1 mole = atomic weight = 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{no}}{\text{.of atoms}}\]
Step by step solution:
Now at first, according to the mole concept we know 1 mole of any substance is equal to its molecular weight. And 1 mole is also equal to the Avogadro’s number of atoms i.e.\[{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\].
Therefore,\[{\text{1 mole = atomic weight = 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{no}}{\text{.of atoms}}\]
According to this it can be said that 39 gm potassium contains \[{\text{6}}{{.023}} \times {\text{1}}{{\text{0}}^{23}}\] number of atoms.
39 gm potassium contains \[{\text{6}}{{.023}} \times {\text{1}}{{\text{0}}^{23}}\] number of atoms.
Now, 1 gram potassium contains \[\dfrac{{{\text{6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}}}{{{\text{39}}}}\] number of atoms
So, 117 grams potassium contains \[\dfrac{{{\text{6}}{{.023 \times 1}}{{\text{0}}^{{{23}}}} \times 117}}{{{\text{39}}}}\] number of atoms.
Therefore, 117 grams potassium contains \[{{3}} \times {\text{6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\] number of atoms.
So, the correct option is D.
Note: The formula of equivalent weight is,
\[{\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{number of equivalent moles}}}}\].
For acids the number of equivalents is equal to the number of H+ present in one molecule. And for bases the number of HO- groups present in one molecule. The equivalent weight of any substance cannot be greater than its molecular weight. Either the value of equivalent weight is equal or less than the molecular weight of that substance.
Formula used: \[{\text{1 mole = atomic weight = 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{no}}{\text{.of atoms}}\]
Step by step solution:
Now at first, according to the mole concept we know 1 mole of any substance is equal to its molecular weight. And 1 mole is also equal to the Avogadro’s number of atoms i.e.\[{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\].
Therefore,\[{\text{1 mole = atomic weight = 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{no}}{\text{.of atoms}}\]
According to this it can be said that 39 gm potassium contains \[{\text{6}}{{.023}} \times {\text{1}}{{\text{0}}^{23}}\] number of atoms.
39 gm potassium contains \[{\text{6}}{{.023}} \times {\text{1}}{{\text{0}}^{23}}\] number of atoms.
Now, 1 gram potassium contains \[\dfrac{{{\text{6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}}}{{{\text{39}}}}\] number of atoms
So, 117 grams potassium contains \[\dfrac{{{\text{6}}{{.023 \times 1}}{{\text{0}}^{{{23}}}} \times 117}}{{{\text{39}}}}\] number of atoms.
Therefore, 117 grams potassium contains \[{{3}} \times {\text{6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\] number of atoms.
So, the correct option is D.
Note: The formula of equivalent weight is,
\[{\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{number of equivalent moles}}}}\].
For acids the number of equivalents is equal to the number of H+ present in one molecule. And for bases the number of HO- groups present in one molecule. The equivalent weight of any substance cannot be greater than its molecular weight. Either the value of equivalent weight is equal or less than the molecular weight of that substance.
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