QUESTION 3 What mass (in g) of glucose, C6H1206, must be added to 300.7 g of water to prepare a 0.322 molal solution? Enter your answer with 3 significant figures and no units.

17.42 g of glucose, C₆H₁₂O₆

Explanation:

From the question given above, the following data were obtained:

Mass of water = 300.7 g

Molar solution = 0.322 M

Mass of glucose =.?

Next, we shall convert 300.7 g to L. This can be obtained as follow:

1000 g = 1 L

Therefore,

300.7 g = 300.7 g × 1 L / 1000 g

300.7 g = 0.3007 L

Next, we shall determine the number of mole of glucose in the molar solution. This can be obtained as follow:

Volume of solution = 0.3007 L

Molar solution = 0.322 M

Mole of glucose =?

Molar solution = mole / Volume

0.322 = mole of glucose / 0.3007

Cross multiply

Mole of glucose = 0.322 × 0.3007

Mole of glucose = 0.0968 mole

Finally, we shall determine the mass of glucose, C₆H₁₂O₆, needed to produce prepare the solution. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.0968 mole

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

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mole = mass /Molar mass

0.0968 = Mass of C₆H₁₂O₆ / 180

Cross multiply

Mass of C₆H₁₂O₆ = 0.0968 × 180

Mass of C₆H₁₂O₆ = 17.42 g

Thus, 17.42 g of glucose, C₆H₁₂O₆ is needed to prepare the molar solution.