Question

18g C6H12O6 (glucose) is dissolved in 180g water. Find the mole fraction of glucose and water.

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Answer 1

Given :

➾ Weight of Glucose = 18g

➾ Weight of Water = 180g

To Find :

➨ Mole fraction of each compound.

Mole Fraction :

➶ Definition : Ratio of number of moles of one component to the total number of moles of all the component.

➶ Effect of temperature : No effect

➶ Symbol : x

➶ Formula : x(A) = n(A)/[n(A)+n(B)]

Calculation :

✭ Moles of Glucose :

➛ Molecular weight = 180g/mol

➛ Mole n(1) = 18/180 = 0.1

✭ Moles of Water :

➛ Molecular weight = 18g/mol

➛ Mole n(2) = 180/18 = 10

➢ Σ(moles) = 10 + 0.1 = 10.1

✭ Mole fraction of Glucose :

⇒ x(1) = n(1) / n(total)

⇒ x(1) = 0.1/10.1

⇒ x(1) = 0.01

✭ Mole fraction of Water :

⇒ x(2) = n(2) / n(total)

⇒ x(2) = 10/10.1

⇒ x(2) = 0.99

Answer 2

$\bigstar$ $\sf{\orange{\underline{\underline{\purple{To\:Find:-}}}}}$

1. Mole fraction of Glucose .
2. Mole fraction of Water .

$\bigstar$ $\sf{\orange{\underline{\underline{\purple{SOLUTION:-}}}}}$

☞ GIVEN:-

• Mass of Glucose taken = 18g
• Mass of Water taken = 180g

(╬) Now, calculate No. of moles of Both

$\tt{\checkmark{\orange{\underline{\:No.\:of\:Moles\:of\:Glucose\::\:}}}}$

$\tt{\:=\:{\dfrac{Given\:Mass}{Molecular\:Mass}}\:}$

$\tt{\:=\:{\dfrac{18}{180}}\:}$

$\tt{\:=\:0.1\:}$

$\tt{\checkmark{\orange{\underline{\:No.\:of\:Moles\:of\:Water\::\:}}}}$

$\tt{\:=\:{\dfrac{180}{18}}\:}$

$\tt{\:=\:10\:}$

$\tt{\checkmark{\orange{\underline{\:Total\:No.\:of\:moles\:of\:solution\::\:}}}}$

$\tt{\:=\:0.1\:+\:10\:}$

$\tt{\:=\:10.1\:}$

$\tt{\checkmark{\orange{\underline{\:Mole\:Fraction\:of\:Glucose\::\:}}}}$

$\tt{\:=\:{\dfrac{0.1}{10.1}}\:}$

$\tt{\:=\:9.9009\times10^{-3}\:}$

$\tt{\boxed{\implies\:Mole\:Fraction\:of\:Glucose\:=\:0.0099009\:}}$

$\tt{\checkmark{\orange{\underline{\:Mole\:Fraction\:of\:Water\::\:}}}}$

$\tt{\:=\:{\dfrac{10}{10.1}}\:}$

$\tt{\boxed{\implies\:Mole\:Fraction\:of\:Water\:=\:0.99009\:}}$