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Answer :
The composition, in atom percent, of an alloy consisting of 4.5 wt% Pb and 95.5 wt% Sn is 2.6 at% Pb and 97.4 at% Sn. The correct answer is (A).
The atomic weights for Pb and Sn are 207.19 g/mol and 118.71 g/mol, respectively.
To determine the atom percent, we need to follow these steps:
- Calculate the moles of Pb and Sn in 100 g of alloy.
Moles of Pb: 4.5 g / 207.19 g/mol = 0.0217 mol
Moles of Sn: 95.5 g / 118.71 g/mol = 0.8047 mol
- Calculate the total moles of atoms in the alloy.
Total moles = 0.0217 mol (Pb) + 0.8047 mol (Sn) = 0.8264 mol
- Calculate the atom percent for each element.
Atom percent of Pb: (0.0217 mol / 0.8264 mol) * 100 ≈ 2.6 at% Pb
Atom percent of Sn: (0.8047 mol / 0.8264 mol) * 100 ≈ 97.4 at% Sn
Thus, the correct answer is (A) 2.6 at% Pb and 97.4 at% Sn.
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Answer: The correct answer is Option A.
Explanation:
We are given:
4.5 wt % of Pb means that 4.5 grams of lead is present in 100 g of alloy.
95.5 wt % of Sn means that 95.5 grams of tin is present in 100 g of alloy.
To calculate the atom percent of any compound in a mixture, we use the equation:
[tex]\text{atom }\%=\frac{\text{Moles of compound}\times N_A}{\text{Total number of moles of mixture}\times N_A}\times 100[/tex]
where,
[tex]N_A[/tex] = Avogadro's number
Moles of a compound is given by the formula:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
- For Lead:
Given mass of lead = 4.5 g
Molar mass of lead = 207.19 g/mol
[tex]\text{Atom percent of lead}=\left(\frac{\frac{4.5g}{207.17g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of lead}=2.6\%[/tex]
- For Tin:
Given mass of tin = 95.5 g
Molar mass of tin = 118.71 g/mol
[tex]\text{Atom percent of Tin}=\left(\frac{\frac{95.5g}{118.71g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of Tin}=97.4\%[/tex]
Hence, the correct answer is Option A.
