What volume of a 0.176 M Na2S solution contains 1.01 g of Na+ ions?
Please show work, I appreciate your help
Sure, no problem! But before I show my work, let me just say that if math problems had faces, this one would definitely have a Na+-ion-sized smile. So, let's dive into it!
First, we need to determine the number of moles of Na+ ions present in 1.01 g. To do this, we can use the molar mass of Na+ ions, which is approximately 22.99 g/mol. So, dividing 1.01 g by the molar mass, we get:
1.01 g / 22.99 g/mol = 0.0439 mol
Now that we have the number of moles of Na+ ions, we can use the formula:
Molarity (M) = Moles (mol) / Volume (L)
Rearranging the formula to solve for volume, we get:
Volume (L) = Moles (mol) / Molarity (M)
Plugging in the values, we have:
Volume (L) = 0.0439 mol / 0.176 M ≈ 0.249 L
Therefore, the volume of the 0.176 M Na2S solution containing 1.01 g of Na+ ions is approximately 0.249 liters. In other words, it's like a tiny drop in an ocean of laughter! Hope that helps!
To find the volume of the Na2S solution containing 1.01 g of Na+ ions, we need to use the given molar concentration (0.176 M) and the molar mass of Na+.
Step 1: Calculate the molar mass of Na+.
The molar mass of Na+ is the same as the molar mass of sodium, which is 22.99 g/mol.
Step 2: Convert the given mass of Na+ ions to moles.
Using the formula: Moles = Mass / Molar mass
Moles = 1.01 g / 22.99 g/mol
Moles = 0.0439 mol
Step 3: Use the molar concentration to find the volume.
The molar concentration tells us the number of moles of Na2S per liter of solution. So, we can use this information to find the volume in liters.
Volume (L) = Moles / Molar concentration
Volume (L) = 0.0439 mol / 0.176 mol/L
Volume (L) = 0.249 L
Step 4: Convert the volume to milliliters (mL) if needed.
Volume (mL) = Volume (L) x 1000
Volume (mL) = 0.249 L x 1000
Volume (mL) = 249 mL
Therefore, the volume of the 0.176 M Na2S solution containing 1.01 g of Na+ ions is 249 mL.
1.01 g Na/molar mass Na = moles Na^+.
moles Na2S = 1/2 moles Na^+.
M Na2S = moles/L.
Solve for L.