How many grams of lithium nitrate, LiNO3 (68.9 g/mol) are required to prepare 219.5 mL of a 0.357 M LiNO3 solution?
The Correct Answer and Explanation is :
To calculate the number of grams of lithium nitrate (LiNO₃) required to prepare 219.5 mL of a 0.357 M LiNO₃ solution, follow these steps:
Step 1: Determine the number of moles required.
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. So, the number of moles of lithium nitrate required can be calculated by the formula:
[
\text{moles of LiNO₃} = Molarity \times Volume \, (\text{in liters})
]
Given:
- Molarity (M) = 0.357 M
- Volume (V) = 219.5 mL = 0.2195 L (since 1 L = 1000 mL)
[
\text{moles of LiNO₃} = 0.357 \, \text{M} \times 0.2195 \, \text{L} = 0.0784 \, \text{moles}
]
Step 2: Convert moles of LiNO₃ to grams.
To find the mass of lithium nitrate required, use the molar mass of LiNO₃. The molar mass of LiNO₃ is given as 68.9 g/mol.
[
\text{mass of LiNO₃} = \text{moles of LiNO₃} \times \text{molar mass of LiNO₃}
]
[
\text{mass of LiNO₃} = 0.0784 \, \text{moles} \times 68.9 \, \text{g/mol} = 5.41 \, \text{g}
]
Final Answer:
To prepare 219.5 mL of a 0.357 M LiNO₃ solution, 5.41 grams of lithium nitrate are required.
Explanation:
Molarity (M) provides the relationship between the volume and the number of moles of solute in a solution. By converting the volume of the solution to liters and using the given molarity, we can find the number of moles of LiNO₃ needed. Then, by using the molar mass of LiNO₃ (68.9 g/mol), we convert the moles into grams, which tells us the exact mass of lithium nitrate required to make the solution.