How many grams of glucose (C6H12O6, 180.2 g/mol) would be required to prepare 2.00 L of a 0.318 M glucose solution that could be used in an IV bag?

115 g

Explanation:

Step 1: Given data

Step 2: Calculate the moles of glucose

The molarity is equal to the moles of solute divided by the liters of solution.

M = moles of glucose / liters of solution

moles of glucose = M × liters of solution

moles of glucose = 0.318 mol/L × 2.00 L

moles of glucose = 0.636 mol

Step 3: Calculate the mass corresponding to 0.636 moles of glucose

0.636 mol × 180.2 g/mol = 115 g

yes

explanation:

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