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Question 6.6: If 5.00 g glucose are dissolved in 1.00 × 10² mL of solution...

If 5.00 g glucose are dissolved in 1.00 × 10² mL of solution, calculate the molarity, M, of the glucose solution.

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Step 1. To use our expression for molarity it is necessary to convert from units of grams of glucose to moles of glucose. The molar mass of glucose is 1.80 × 10² g/mol. Therefore

5.00 g×1 mol1.80×102 g=2.78×10−2 mol glucose 5.00  \cancel{g} \times \frac{1  mol}{1.80 \times 10^2  \cancel{g}} = 2.78 \times 10^{-2}   mol  glucose

Step 2. We must convert mL to L:

1.00×102 mL×1 L103 mL=1.00×10−1 L1.00 \times 10^2  \cancel{mL} \times \frac{1  L}{ 10^3  \cancel{mL}} = 1.00 \times 10^{-1}   L

Step 3. Substituting these quantities:

Mglucose=2.78×10−2 mol1.00×10−1 L= 2.78×10−1 MM_{glucose}= \frac{2.78 \times 10^{-2}   mol}{1.00 \times 10^{-1}   L} \\ =  2.78 \times 10^{-1}  M

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