Answer:
17.42 g of glucose, C₆H₁₂O₆
Step-by-step 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.