Question

20 % (w/w) aqueous solution of glucose is provided. The mole fraction of glucose in this solution is
A) 0.024
B) 0.036
C) 0.975
D) 0.858​

Answer 1

Answer:

0.975. please mark me as branalist

Answer 2

Answer:

The glucose's mole fraction in the aqueous solution calculated is $0.024$.

Therefore, option a) $0.024$ is correct.

Explanation:

Given data,

The concentration ( W/W% ) of the aqueous solution of glucose = $20\%$

The mole fraction of the glucose in the aqueous solution =?

As we know,

• Mole fraction = $\frac{n_{solute} }{n_{solute} + n_{solvent} }$

So, in the given case;

• Mole fraction = $\frac{n_{glucose} }{n_{glucose} + n_{water} }$

Here,

• $n_{glucose}$ = the number of moles of glucose
• $n_{solute}$ = the number of moles of the solute ( water )

Now, as given;

• W/W% = $20\%$ = $\frac{20}{100}$

This means the mass of glucose = $20g$

And the mass of the solution = $100g$

Therefore, the mass of water = $100-20$ = $80g$

Now, we have to calculate the number of moles of solute and solution:

• The molar mass of glucose = $180g/mol$
• The molar mass of water = $18g/mol$

Thus,

• The number of moles of glucose = $\frac{ mass}{molar mass}$ = $\frac{20}{180}$ = $0.11mol$
• The number of moles of water = $\frac{ mass}{molar mass}$ = $\frac{80}{18}$ = $4.44mol$

Therefore,

• Mole fraction = $\frac{n_{glucose} }{n_{glucose} + n_{water} }$
• Mole fraction = $\frac{0.11}{0.11+4.44}$
• Mole fraction = $\frac{0.11}{4.55}$
• Mole fraction = $0.024$

Hence, the mole fraction of the glucose in the aqueous solution = $0.024$.