Question

What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn? The atomic weights for Pb and Sn are 207.19 g/mol and 118.71 g/mol, respectively.(A) 2.6 at% Pb and 97.4 at% Sn(B) 7.6 at% Pb and 92.4 at% Sn(C)97.4 at% Pb and 2.6 at% Sn(D) 92.4 at% Pb and 7.6 at% Sn

Answer 1

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:

$\text{atom }\%=\frac{\text{Moles of compound}\times N_A}{\text{Total number of moles of mixture}\times N_A}\times 100$

where,

$N_A$ = Avogadro's number

Moles of a compound is given by the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

• For Lead:

Given mass of lead = 4.5 g

Molar mass of lead = 207.19 g/mol

$\text{Atom percent of lead}=\left(\frac{\frac{4.5g}{207.17g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of lead}=2.6\%$

• For Tin:

Given mass of tin = 95.5 g

Molar mass of tin = 118.71 g/mol

$\text{Atom percent of Tin}=\left(\frac{\frac{95.5g}{118.71g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of Tin}=97.4\%$

Hence, the correct answer is Option A.