What is the molality of a glucose solution prepared by dissolving 16.7 g of glucose, C6H12O6, in 133.6 g of water

0.696 m

Explanation:

We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:

Mass of C₆H₁₂O₆ = 16.7 g

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)

= 72 + 12 + 96

= 180 g/mol

Mole of C₆H₁₂O₆ =?

Mole = mass / molar mass

Mole of C₆H₁₂O₆ = 16.7 / 180

Mole of C₆H₁₂O₆ = 0.093 mole

Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

133.6 g = 133.6 g × 1 Kg / 1000 g

133.6 g = 0.1336 Kg

Thus, 133.6 g is equivalent to 0.1336 Kg.

Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:

Mole of C₆H₁₂O₆ = 0.093 mole

Mass of water = 0.1336 Kg

Molality =?

Molality = mole / mass of water (in Kg)

Molality = 0.093 / 0.1336

Molality = 0.696 m

Therefore, the molality of the solution is 0.696 m

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.

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