Chapter 13: Problem 58

A package of compounds used to achieve rehydration in sick patients contains 20.0 g of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) . When this material is diluted to \(1.00 \mathrm{L},\) what is the molarity of glucose?

Short Answer

Expert verified

The molarity of glucose in the solution is approximately 0.111 M.

Step by step solution

Calculate the molar mass of glucose

First, calculate the molar mass of glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)). The molar mass of an element is its atomic mass expressed in grams. Using the atomic masses of carbon (C=12.01g/mol), hydrogen (H=1.008g/mol), and oxygen (O=16.00g/mol) from the periodic table, we find the molar mass of glucose as follows:\[\mathrm{Molar~mass~of~glucose} = 6(12.01g/mol)+12(1.008g/mol)+6(16.00g/mol) = 180.16g/mol\]

Convert grams of glucose to moles

Next, convert the mass of glucose into moles. We know that 20.0g of glucose is used and the molar mass of glucose is 180.16g/mol. Using the formula, moles = mass / molar mass, we find the moles as follows:\[\mathrm{Moles~of~glucose} = \frac{20.0g}{180.16g/mol} \approx 0.111 ~moles\]

Calculate the molarity of glucose

After finding the moles, the next step is to determine the molarity. The molarity (M) can be calculated by dividing the moles of solute by the liters of solution. With a volume of 1.00L of solution, we can calculate the molarity as follows:\[\mathrm{Molarity~of~glucose} = \frac{0.111 ~moles}{1.00 L} \approx 0.111 M\]

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A package of compounds used to achieve rehydration in sick patients contains 20.0 g of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) . When this material is diluted to \(1.00 \mathrm{L},\) what is the molarity of glucose?

Short Answer

Step by step solution

Calculate the molar mass of glucose

Convert grams of glucose to moles

Calculate the molarity of glucose

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