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✨A solution of glucose in water is labelled as 10% (weight /weight) what will be the molality and mole fraction of each component in the solution . If the density of the solution is 1.2 gram per ml then what shall be the molarity of the solution?

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Answer 1

10%w/w solution of glucose in water means 10 g of glucose present in 100g of solution.

Therefore,

Mass of glucose is 10g Mass of water is = 100g- 10g = 90 g = 0.09 kg mAnd density = 1.2gmL-1 .

Let us find molality Molality = Number of moles of solute / Mass of solvent in Kg

No of moles of solute = Mass of glucose / Molar mass of glucose Molar mass of glucose (C6H12O6)

= 6x12 + 12x1 + 6x16 =180 g /mol Number of moles of glucose =

10/180 = 0.0556 moles

Molality = 0.0556/0.09 = 16.7M

Now find the molarity ,

Molarity = Number of moles of solute / Volume of solution in liter Density = 1.2 g/ml

Volume = mass / density = 100 g / 1.2 g/ ml = 83.33 ml = 0.0833 L

Molarity = 0.0556/ 0.0833 = 0.67 M

Answer 2

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❎10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.

⭐Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1

⭐Then, number of moles of glucose = 10 / 180 mol

= 0.056 mol

∴ Molality of solution = 0.056 mol / 0.09kg  = 0.62 m

✔Number of moles of water = 90g / 18g mol-1 = 5 mol

✔Mole fraction of glucose  (xg) = 0.056 / ( 0.056+5) = 0.011

✔And, mole fraction of water xw = 1 - xg

= 1 - 0.011

= 0.989

⭐If the density of the solution is 1.2 g mL - 1, then the volume of the 100 g solution can be given as:

= 100g / 1.2g mL-1

= 83.33 mL

=83.33 x 10-3 L

∴ Molarity of the solution = 0.056 mol / 83.33 x 10-3 L

= 0.67 M