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Chapter 12: Problem 108

How many milliliters of a \(0.250 \mathrm{M}\) solution of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), are required to obtain \(100.0 \mathrm{~g}\) of glucose?

Short Answer

Expert verified
To obtain 100.0 g of glucose, 2220 mL of a 0.250 M glucose solution is required.

Step by step solution

01

Calculate the molecular weight of glucose

First, we need to find the molecular weight of glucose. The formula of glucose is C6H12O6, and the atomic weights of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively. Molecular weight of glucose = (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol
02

Convert grams of glucose into moles

Now, we need to convert the given mass of glucose (100.0 g) into moles. To do this, we will use the molecular weight of glucose (180.18 g/mol) that we calculated in the previous step. Moles of glucose = 100.0 g / 180.18 g/mol = 0.555 moles
03

Find the volume of the solution needed

Now that we have the moles of glucose, we can use the given molarity and the formula for molarity to find the volume of the solution in liters. Molarity formula: M = moles of solute / volume of solution in liters 0.250 M = 0.555 moles / volume of solution in liters Volume of solution in liters = 0.555 moles / 0.250 M = 2.22 L
04

Convert the volume from liters to milliliters

Finally, we need to convert the volume of the solution from liters to milliliters. 1 L = 1000 mL Volume of solution in milliliters = 2.22 L × 1000 mL/L = 2220 mL So, 2220 mL of a 0.250 M solution of glucose is required to obtain 100.0 g of glucose.

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