How many grames of lithium nitrate LiNO3 (68.9 g/mol) are required to prepare 174.1 mL of a 0.698 MLiNO3 solution ?

Same procedure as in those other similar problems.

To determine the number of grams of lithium nitrate (LiNO3) required to prepare a specific volume and concentration of solution, you'll need to use the formula:

Amount (in grams) = concentration (in moles per liter) × volume (in liters) × molar mass (in grams per mole)

Let's break down the given information and solve the problem step by step:

1. Calculate the number of moles of LiNO3 needed in the solution:
moles = concentration × volume
Since the given concentration is in millimoles per liter (mM), we need to convert it to moles per liter (M):
0.698 mM LiNO3 = 0.698 × 10^-3 mol/L LiNO3
Note: millimoles (mM) are equal to moles (mol) divided by 1000.

2. Convert the volume of the solution to liters:
Given volume = 174.1 mL = 174.1 × 10^-3 L

3. Calculate the molar mass of LiNO3:
The molar mass of lithium nitrate (LiNO3) is given as 68.9 g/mol.

Now we can combine all the values to calculate the amount of LiNO3 needed:

Amount (in grams) = concentration (in moles per liter) × volume (in liters) × molar mass (in grams per mole)
= (0.698 × 10^-3 mol/L) × (174.1 × 10^-3 L) × 68.9 g/mol

By performing the calculation, you will find the number of grams of LiNO3 required to prepare the given solution.