What volume, in liters, of $1.0 M C_6 H_{12} O_6$ solution can be made using $360.4 g C_6 H_{12} O_6$?
$C_6 H_{12} O_6: 180.2 g / mol$
Answer (2)
You can prepare 2.0 liters of a 1.0 M glucose solution using 360.4 grams of glucose. This is calculated by determining the number of moles of glucose and using the molarity formula. Therefore, the answer is 2.0 liters.
;
Calculate the number of moles of C 6 H 12 O 6 : m o l es = 180.2 g/mol 360.4 g = 2.0 mol .
Use the molarity formula to find the volume: v o l u m e = M o l a r i t y m o l es .
Substitute the values: v o l u m e = 1.0 M 2.0 mol = 2.0 L .
The volume of the solution is 2.0 L .
Explanation
Problem Analysis We are given the mass of C 6 H 12 O 6 (360.4 g), the molar mass of C 6 H 12 O 6 (180.2 g/mol), and the molarity of the solution (1.0 M). We need to find the volume of the solution in liters.
Calculating Moles First, we need to calculate the number of moles of C 6 H 12 O 6 using the formula: m o l es = m o l a r _ ma ss ma ss Substituting the given values: m o l es = 180.2 g/mol 360.4 g = 2.0 mol So, we have 2.0 moles of C 6 H 12 O 6 .
Calculating Volume Next, we use the definition of molarity: M o l a r i t y = v o l u m e ( in L ) m o l es We are given the molarity (1.0 M) and we have calculated the number of moles (2.0 mol). We need to find the volume. Rearranging the formula to solve for volume: v o l u m e ( in L ) = M o l a r i t y m o l es Substituting the values: v o l u m e ( in L ) = 1.0 M 2.0 mol = 2.0 L Therefore, the volume of the solution is 2.0 liters.
Final Answer The volume of the 1.0 M C 6 H 12 O 6 solution that can be made using 360.4 g of C 6 H 12 O 6 is 2.0 liters.
Examples
In a lab setting, you might need to prepare solutions of specific concentrations for experiments. For example, if you need a 1.0 M glucose solution for a cell culture experiment, you would use the same calculation to determine how much glucose to dissolve in a certain volume of water. This ensures that your cells receive the correct concentration of nutrients. Knowing how to calculate molarity and volume is crucial in many scientific applications, from preparing reagents to conducting research.
