A. For a two-component alloy, derive the equations that convert the weight percent to atomic percent (for a two-element alloy). b. Calculate the atomic fraction of Cu in an alloy that consists of 92.5 wt.% Ag and 7.5wt% Cu

a. wt% A / Atomic Mass A / (wt% A / Atomic Mass A + wt% B / Atomic Mass B).

b. 11.74%

Explanation:

a. Weight percent is the mass of an element in 100g of the alloy. Atomic percent could be taken as the ratio between moles of an element and the total moles of the alloy.

Assuming a basis of 100g, the moles of A and B (The two elements in the alloy), are:

Moles A:

wt% A / Atomic Mass A

Moles B:

wt% B / Atomic Mass B

And atomic percent of A is:

b. Atomic fraction of Cu in the alloy is (Molar mass Cu: 63.55g/mol; Molar mass Ag: 107.87g/mol):

7.25wt% / 63.55g/mol / (7.25wt% / 63.55g/mol + 92.5wt% / 107.87g/mol)

Atomic fraction Cu = 0.1174 = 11.74%

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.