What is the molarity of an aqueous solution containing 56.2 g of glucose (C6H12O6) in 2.3 L of solution?

4.90 mol L

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1

Explanation:

Start by calculating the number of moles of glucose present in  

25.5 mL

of solution.

To do that, use the molar mass of glucose

22.5

g

⋅

1 mole glucose

180.156

g

=

0.1249 moles glucose

Now, in order to find the molarity of the solution, you need to determine the number of moles of solute present in  

1 L

=

10

3

mL

of solution.

The fact that you're dealing with a solution, which, as you know, is a heterogeneous mixture, i.e. it has the same composition throughout, you can use the known composition as a conversion factor to find the number of moles of solute present in  

10

3

mL

of solution.

mL solution

⋅

0.1249 moles glucose

25.5

mL solution

=

4.898 moles glucose

You can thus say that the molarity of the solution is equal to

molarity = 4.90 mol L

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1

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0

explanation:

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