Answer:
2.70 g of glucose.
Explanation:
The following data were obtained from the question:
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Molar mass of glucose = 180.16 g/mol
Mass of glucose =.?
Next, we shall determine the number of mole of glucose in the solution. This can be obtained as follow:
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Mole glucose =?
Molarity = mole /Volume
0.1 = Mole of glucose /0.15
Cross multiply
Mole of glucose = 0.1 × 0.15
Mole of glucose = 0.015 mole
Finally, we shall determine the mass of glucose in the solution as follow:
Mole of glucose = 0.015 mole
Molar mass of glucose = 180.16 g/mol
Mass of glucose =.?
Mole = mass /molar mass
0.015 = mass of glucose /180.16
Cross multiply
Mass of glucose = 0.015 × 180.16
Mass of glucose = 2.70 g
Therefore, the solution contains 2.70 g of glucose.
The grams that the solution must contain is :
- 2.70 g of glucose.
Given:
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Molar mass of glucose = 180.16 g/mole
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Molarity = mole /Volume
0.1 = Mole of glucose /0.15
Mole of glucose = 0.1 × 0.15
Mole of glucose = 0.015 mole
Mole of glucose = 0.015 mole
Molar mass of glucose = 180.16 g/mol
Mass of glucose =.?
Mole = mass /molar mass
0.015 = mass of glucose /180.16
Mass of glucose = 0.015 × 180.16
Mass of glucose = 2.70 g
The solution contains 2.70 g of glucose.
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