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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

$\huge\bold\pink{Answer:-}$

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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✔✔✔

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Answer 2

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