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Stoichiometry - Engineering

Q1:

The weight fraction of methanol in an aqueous solution is 0.64. The mole fraction of methanol XMsatisfies

A XM < 0.5

B XM = 0.5

C 0.5 < XM < 0.64

D XM ≥ 0.64

ANS:D - XM ≥ 0.64

To relate weight fraction (𝑊𝑀WM​) and mole fraction (𝑋𝑀XM​), we can use the following relationship: 𝑊𝑀=𝑋𝑀⋅𝑀𝑀𝑋𝑀⋅𝑀𝑀+𝑋𝑊⋅𝑀𝑊WM​=XM​⋅MM​+XW​⋅MW​XM​⋅MM​​ Where:

  • 𝑊𝑀WM​ = weight fraction of methanol (given as 0.64)
  • 𝑋𝑀XM​ = mole fraction of methanol
  • 𝑋𝑊XW​ = mole fraction of water
  • 𝑀𝑀MM​ = molar mass of methanol
  • 𝑀𝑊MW​ = molar mass of water
Given that it's an aqueous solution, the mole fraction of water (𝑋𝑊XW​) can be calculated as 1−𝑋𝑀1−XM​. The molar masses of methanol (𝑀𝑀MM​) and water (𝑀𝑊MW​) are constants. Let's calculate 𝑋𝑀XM​: 0.64=𝑋𝑀⋅𝑀𝑀𝑋𝑀⋅𝑀𝑀+(1−𝑋𝑀)⋅𝑀𝑊0.64=XM​⋅MM​+(1−XM​)⋅MW​XM​⋅MM​​ Since 𝑀𝑀MM​ and 𝑀𝑊MW​ are constants, we can consider their ratio as 𝑘k for simplicity: 0.64=𝑋𝑀𝑋𝑀+(1−𝑋𝑀)⋅𝑘0.64=XM​+(1−XM​)⋅kXM​​ Now, let's solve for 𝑋𝑀XM​: 0.64=𝑋𝑀1−𝑋𝑀+𝑋𝑀⋅𝑘0.64=1−XM​+XM​⋅kXM​​ 0.64(1−𝑋𝑀+𝑋𝑀⋅𝑘)=𝑋𝑀0.64(1−XM​+XM​⋅k)=XM​ 0.64−0.64𝑋𝑀+0.64𝑋𝑀⋅𝑘=𝑋𝑀0.64−0.64XM​+0.64XM​⋅k=XM​ 0.64=𝑋𝑀(1+0.64𝑘−0.64)0.64=XM​(1+0.64k−0.64) 𝑋𝑀=0.640.64𝑘XM​=0.64k0.64​ As 𝑘k is greater than 1 (since the molar mass of water is greater than methanol), the value of 0.640.64𝑘0.64k0.64​ is less than 1. Therefore, 𝑋𝑀<0.5XM​<0.5. So, the correct option is 𝑋𝑀<0.5XM​<0.5.