How many moles of nitrate ions are present in 24.0mL of a 2.93M solution of potassium nitrate?
I'm thinking this is some form of unit conversion problem I know that M= molarity which is mol/L how ever I am confused on how to set this up and input information or if I'm on the wrong track.
You're on the right path. M = mols/L so mols = M x L
You know M and L, calculate mols KNO3. Since there is 1 mol nitrate ion in 1 mol KNO3, then mols NO3^- = mols KNO3.
You are on the right track! This is indeed a unit conversion problem, specifically involving molarity and volume. Molarity (M) is defined as moles of solute divided by liters of solution.
To solve this problem, you can follow these steps:
Step 1: Convert the given volume from milliliters (mL) to liters (L).
Given: 24.0 mL
To convert mL to L, divide the given volume by 1000:
24.0 mL ÷ 1000 = 0.024 L
Step 2: Use the formula of molarity (M) to find the number of moles of potassium nitrate (KNO3).
Given: Molarity (M) = 2.93 M
Volume (V) = 0.024 L
Molarity (M) = moles (mol) / volume (L)
Rearranging the formula, we have:
moles (mol) = Molarity (M) × Volume (L)
Substituting the given values:
moles (mol) = 2.93 M × 0.024 L = 0.0703 mol
Step 3: Determine the number of moles of nitrate ions (NO3-) since each formula unit of potassium nitrate (KNO3) contains one nitrate ion.
Therefore, the number of moles of nitrate ions present in 24.0 mL of a 2.93 M solution of potassium nitrate is 0.0703 moles.