Question

92 g of ethyl alcohol (C2H5OH), 62 g of ethylene glycol
(C2H602) and 96 g of methyl alcohol (CH3OH) are
present in a solution. The mole fraction of ethylene
glycol in the solution is
0.17
0.33
0.5
0.25​

Answer 1

Explanation:

92g of ethyl alcohol (C2H5OH), 62g of ethylene glycol (C2H602) and 96g of methyl alcohol (CH3OH) are present in a solution. The mole fraction of ethylene

glycol in the solution is

a) 0.17 ✔

b) 0.33

c) 0.5

d) 0.25

Explaination :-

Let's calculate the molecular mass of All compound :-

1) C2H5OH

→ 12×2 + 1×6 + 16

→ 24 + 6 + 16

→ 46 g/mol

2) C2H6O2

→ 12×2 + 1×6 + 16×2

→ 24 + 6 + 32

→ 62 g/mol

3) CH3OH

→ 12 + 1×4 + 16

→ 12 + 4 + 16

→ 32 g/mol

Now,

Let's calculate the mole of all compound :-

1) C2H5OH

→ 92/46

→ 2 mol

2) C2H6O2

→ 62/62

→ 1 mol

3) CH3OH

→ 96/32

→ 3 mol

Now,

As we know that :-

mole fraction of a component =

• number of mole of the component ÷ total number of moles of all the components

→ 1/6

→ 0.166

≈ 0.17

Therefore,

Option A is correct.