How many grams of lithium nitrate, LiNO3 (68.9 g/mol), are required to prepare 781.0mL of a 0.100 M LiNO3 solution?
The Correct Answer and Explanation is :
To prepare 781.0 mL of a 0.100 M lithium nitrate (LiNO₃) solution, you need to determine the mass of LiNO₃ required. Here’s how to calculate it:
Step 1: Convert volume to liters
Since molarity (M) is expressed in moles per liter, first convert the volume from milliliters to liters:
781.0 mL × (1 L / 1000 mL) = 0.781 L
Step 2: Calculate moles of LiNO₃ needed
Use the molarity formula:
Moles = Molarity × Volume
Moles = 0.100 mol/L × 0.781 L = 0.0781 mol
Step 3: Calculate mass of LiNO₃ required
The molar mass of LiNO₃ is 68.9 g/mol. To find the mass:
Mass = Moles × Molar Mass
Mass = 0.0781 mol × 68.9 g/mol = 5.38109 g
Therefore, approximately 5.38 grams of LiNO₃ are needed.
Explanation:
Molarity (M) represents the concentration of a solution, defined as the number of moles of solute per liter of solution. To prepare a solution with a specific molarity and volume, you can calculate the required moles of solute by multiplying the molarity by the volume in liters. Once you have the moles, you can convert this to grams using the molar mass of the solute.
In this case, the desired concentration is 0.100 M, and the volume is 781.0 mL (which converts to 0.781 L). Multiplying these gives the moles of LiNO₃ needed. Then, multiplying the moles by the molar mass of LiNO₃ (68.9 g/mol) provides the mass required.
This method ensures that the solution has the correct concentration for its intended use.