Question

A solution prepared by adding 360 g of glucose to 864 g of water. calculate mole fraction of glucose. (molar mass of glucose = 180​

Answer 1

Given mass of water, $\sf{W_A=864\ g}$

Given mass of glucose, $\sf{W_B=360\ g}$

Molar mass of water, $\sf{M_A=18\ g\,mol^{-1}}$

Molar mass of glucose, $\sf{M_B=180\ g\,mol^{-1}}$

No. of moles of water,

$\sf{\longrightarrow n_A=\dfrac{W_A}{M_A}}$

$\sf{\longrightarrow n_A=\dfrac{864\ g}{18\ g\,mol^{-1}}}$

$\sf{\longrightarrow n_A=48\ mol}$

No. of moles of glucose,

$\sf{\longrightarrow n_B=\dfrac{W_B}{M_B}}$

$\sf{\longrightarrow n_B=\dfrac{360\ g}{180\ g\,mol^{-1}}}$

$\sf{\longrightarrow n_B=2\ mol}$

Then, mole fraction of glucose,

$\sf{\longrightarrow \chi_B=\dfrac{n_B}{n_A+n_B}}$

$\sf{\longrightarrow \chi_B=\dfrac{2\ mol}{48\ mol+2\ mol}}$

$\sf{\longrightarrow \chi_B=\dfrac{2\ mol}{50\ mol}}$

$\sf{\longrightarrow\underline{\underline{\chi_B=0.04}}}$

Hence 0.04 is the answer.

Answer 2

Answer:

Given mass of water, \sf{W_A=864\ g}W

A

=864 g

Given mass of glucose, \sf{W_B=360\ g}W

B

=360 g

Molar mass of water, \sf{M_A=18\ g\,mol^{-1}}M

A

=18 gmol

−1

Molar mass of glucose, \sf{M_B=180\ g\,mol^{-1}}M

B

=180 gmol

−1

No. of moles of water,

\sf{\longrightarrow n_A=\dfrac{W_A}{M_A}}⟶n

A

=

M

A

W

A

\sf{\longrightarrow n_A=\dfrac{864\ g}{18\ g\,mol^{-1}}}⟶n

A

=

18 gmol

−1

864 g

\sf{\longrightarrow n_A=48\ mol}⟶n

A

=48 mol

No. of moles of glucose,

\sf{\longrightarrow n_B=\dfrac{W_B}{M_B}}⟶n

B

=

M

B

W

B

\sf{\longrightarrow n_B=\dfrac{360\ g}{180\ g\,mol^{-1}}}⟶n

B

=

180 gmol

−1

360 g

\sf{\longrightarrow n_B=2\ mol}⟶n

B

=2 mol

Then, mole fraction of glucose,

\sf{\longrightarrow \chi_B=\dfrac{n_B}{n_A+n_B}}⟶χ

B

=

n

A

+n

B

n

B

\sf{\longrightarrow \chi_B=\dfrac{2\ mol}{48\ mol+2\ mol}}⟶χ

B

=

48 mol+2 mol

2 mol

\sf{\longrightarrow \chi_B=\dfrac{2\ mol}{50\ mol}}⟶χ

B

=

50 mol

2 mol

\sf{\longrightarrow\underline{\underline{\chi_B=0.04}}}⟶

χ

B

=0.04

Hence 0.04 is the answer.