
Bell metal contains $80\,\% $ copper. The mass of Bell metal which contains $1.5 \times {10^{20}}$ atoms of copper is $\left( {Cu = 64} \right)$.
(A) \[2\,\,mg\].
(B) $20\,\,mg$.
(C) $40\,\,mg$.
Answer
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Hint: Avogadro's constant is a number of units in one mole of any of the substance that is it is explained as its molecular weight in grams, which is equal to $6.022 \times {10^{23}}$. The units may be electrons, atoms, ions, or molecules, which depends on the nature of the substance and the character of the reaction.
Formula used:
Mass of the bell metal is found by;
$m = M \times {M_m}$
Where, $m$ denotes the mass of the bell metal, $M$ denotes the number of moles present in the bell metal, ${M_m}$ denotes the molar mass of the bell metal.
Complete step by step answer:
The data given in the problem are as follows;
Number of atoms of copper is $1.5 \times {10^{20}}$.
Molar mass of the copper is, $64\,\,g\,mo{l^{ - 1}}$.
Percentage of copper present in bell metal is, $80\,\% $.
To find the number of moles in the bell;
According to Avogadro’s constant, we know that $1$ mole of any substance has exactly $6.022 \times {10^{23}}$ atoms or molecules in it;
That is;
$6.022 \times {10^{23}}$ atoms $ = 1\,\,mole$
Then for $1.5 \times {10^{20}}$ of atoms;
$
M = 1 \times \dfrac{{1.5 \times {{10}^{20}}}}{{6.022 \times {{10}^{23}}}} \\
\implies M = 0.00025\,\,mol \\
$
Mass of the bell metal is found by;
$m = M \times {M_m}$
Substitute the values of number of moles in the bell metal and the molar mass of the copper bell metal in the above equation;
$
m = 0.00025\,\,mol \times 64\,\,g\,mo{l^{ - 1}} \\
\implies m = 0.00025 \times 64 \\
\implies m = 0.016\,\,g \\
$
By calculating the percentage of the copper with the total weight percentage;
$
m = \dfrac{{80}}{{100}} \times 0.016 \\
\implies m = 0.02\,\,g \\
$
Converting the found mass from gram into milligram;
$m = 20\,\,mg$
Therefore, the piece of metal of the bell consists of $m = 20\,\,mg$.
So, the correct answer is “Option B”.
Note:
The molar mass of any of the chemical compounds is detailed as the mass of a sample of that compound which is divided by the amount of substance in that particular sample, which is measured in moles.
The molar mass is a volume and not a molecular property of a substance.
Formula used:
Mass of the bell metal is found by;
$m = M \times {M_m}$
Where, $m$ denotes the mass of the bell metal, $M$ denotes the number of moles present in the bell metal, ${M_m}$ denotes the molar mass of the bell metal.
Complete step by step answer:
The data given in the problem are as follows;
Number of atoms of copper is $1.5 \times {10^{20}}$.
Molar mass of the copper is, $64\,\,g\,mo{l^{ - 1}}$.
Percentage of copper present in bell metal is, $80\,\% $.
To find the number of moles in the bell;
According to Avogadro’s constant, we know that $1$ mole of any substance has exactly $6.022 \times {10^{23}}$ atoms or molecules in it;
That is;
$6.022 \times {10^{23}}$ atoms $ = 1\,\,mole$
Then for $1.5 \times {10^{20}}$ of atoms;
$
M = 1 \times \dfrac{{1.5 \times {{10}^{20}}}}{{6.022 \times {{10}^{23}}}} \\
\implies M = 0.00025\,\,mol \\
$
Mass of the bell metal is found by;
$m = M \times {M_m}$
Substitute the values of number of moles in the bell metal and the molar mass of the copper bell metal in the above equation;
$
m = 0.00025\,\,mol \times 64\,\,g\,mo{l^{ - 1}} \\
\implies m = 0.00025 \times 64 \\
\implies m = 0.016\,\,g \\
$
By calculating the percentage of the copper with the total weight percentage;
$
m = \dfrac{{80}}{{100}} \times 0.016 \\
\implies m = 0.02\,\,g \\
$
Converting the found mass from gram into milligram;
$m = 20\,\,mg$
Therefore, the piece of metal of the bell consists of $m = 20\,\,mg$.
So, the correct answer is “Option B”.
Note:
The molar mass of any of the chemical compounds is detailed as the mass of a sample of that compound which is divided by the amount of substance in that particular sample, which is measured in moles.
The molar mass is a volume and not a molecular property of a substance.
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