Question

2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water.


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Answer 1

Answer:-

• Mole fraction of glucose = 0.0095

• Mole fraction of water = 0.9905

Explanation:-

Here, glucose is solute and water

is solvent.

• Molar mass of Carbon (C) = 12g/mol

• Molar mass of Hydrogen (H) = 1g/mol

• Molar mass of Oxygen (O) = 16g/mol

Hence, molar mass of glucose [C₆H₁₂O₆] :-

= 12×6 + 1×12 + 16×6

= 72 + 12 + 96

= 180g/mol

And, molar mass of water [H₂O] :-

= 1×2 + 16

= 18g/mol

Number of mole in 2.82g of glucose :-

= Given Mass/Molar mass

= 2.82/180

= 0.016 mole

Number of mole in 30g of water :-

= Given Mass/Molar mass

= 30/18

= 1.67 mole

Mole fraction of solute [glucose] :-

= Mole of solute/Total mole in solution

= 0.0156/[0.016 + 1.67]

= 0.0156/1.686

= 0.0095

Mole fraction of solvent [water] :-

= Mole of solvent/Total mole in solution

= 1.67/[0.016 + 1.67]

= 1.67/1.686

= 0.9905

Answer 2

Explanation:

We know that,

The weight of glucose=2.82g

Also,

no. of moles of glucose=n2 = 2.82/180 =0.01567

no. of moles of N2O=n1= 30/18 =1.667

Molality = n2/n1 × 1000/m.wt of solvent

= 0.01567/1.667 × 1000/18 =0.5222molal

We have moles of glucose (x2)=0.01567 moles of H2O(x1 )=1.667

So, mole fraction of glucose

x2 = n2/n1 + n2 = 0.0157/(1.667+0.01567) =0.0093

mole fraction of H2O=x1

x1 =1−x2 =1=0.0093

x1 =0.9907

Hope this is helpful for you.