Question

12. Molarity of aqueous solution of glucose is 2.315
Mol/L. Mole fraction of glucose is
(1) 0.05
(2) 0.04
(3) 0.03
(4) 0.02​

Answer 1

Given:

Molarity of aqueous solution of glucose is 2.315 mol/L.

To find:

Mole fraction of glucose ?

Calculation:

Molarity is 2.315 means that the solution has 2.315 moles of glucose in 1000 mL (1 litre) of water.

So, moles of water :

$\therefore \: n_{water} = \dfrac{given \: mass \: }{molar \: mass}$

$\implies\: n_{water} = \dfrac{1000 }{18}$

$\implies\: n_{water} = 55.55 \: mol$

Now, let's find the mole fraction:

$\therefore \: \chi_{(glucose)} = \dfrac{n_{(glucose)}}{n_{(glucose) } + n_{(water)}}$

$\implies \: \chi_{(glucose)} = \dfrac{2.315}{2.315+ 55.55}$

$\implies \: \chi_{(glucose)} = \dfrac{2.315}{57.865}$

$\implies \: \chi_{(glucose)} = 0.04$

So, mole fraction of glucose is 0.04.

Answer 2

Answer:

Given:

Molarity of aqueous solution of glucose is 2.315 mol/L.

To find:

Mole fraction of glucose ?

Calculation:

Molarity is 2.315 means that the solution has 2.315 moles of glucose in 1000 mL (1 litre) of water.

So, moles of water :

\therefore \: n_{water} = \dfrac{given \: mass \: }{molar \: mass}∴n

water

=

molarmass

givenmass

\implies\: n_{water} = \dfrac{1000 }{18}⟹n

water

=

18

1000

\implies\: n_{water} = 55.55 \: mol⟹n

water

=55.55mol

Now, let's find the mole fraction:

\therefore \: \chi_{(glucose)} = \dfrac{n_{(glucose)}}{n_{(glucose) } + n_{(water)}}∴χ

(glucose)

=

n

(glucose)

+n

(water)

n

(glucose)

\implies \: \chi_{(glucose)} = \dfrac{2.315}{2.315+ 55.55}⟹χ

(glucose)

=

2.315+55.55

2.315

\implies \: \chi_{(glucose)} = \dfrac{2.315}{57.865}⟹χ

(glucose)

=

57.865

2.315

\implies \: \chi_{(glucose)} = 0.04⟹χ

(glucose)

=0.04

So, mole fraction of glucose is 0.04.