how many grams of sodium hydroxide, NaOH, are present in 1.50*10^24 molecules of sodium hydroxide?
Is there a formula to solve this?
yes, and clear thinking,
How many moles do you have in 1.5E24 molecules. You know 1 mol of molecules contains 6.02E23 molecules; therefore,
mols of NaOH = 1.5E24/6.023E23 = ? moles.
Now, grams = mols x molar mass = ?
Post your work if you get stuck.
To determine the number of grams of sodium hydroxide (NaOH) present in 1.50 * 10^24 molecules of NaOH, we need to use the concept of molar mass and Avogadro's number.
Step 1: Find the molar mass of NaOH.
The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). The atomic masses of Na, O, and H are 22.99 g/mol, 16.00 g/mol, and 1.01 g/mol, respectively.
So, the molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol.
Step 2: Use Avogadro's number to determine the number of moles of NaOH in 1.50 * 10^24 molecules.
Avogadro's number (6.022 * 10^23) represents the number of particles (atoms, molecules, etc.) in one mole of a substance.
Number of moles = Number of molecules / Avogadro's number
Number of moles = 1.50 * 10^24 molecules / 6.022 * 10^23 molecules/mol
Number of moles = 2.49 moles
Step 3: Calculate the mass of NaOH using the molar mass and the number of moles.
Mass (grams) = Number of moles * Molar mass
Mass (grams) = 2.49 moles * 40.00 g/mol
Mass (grams) ≈ 99.6 grams
Therefore, there are approximately 99.6 grams of sodium hydroxide (NaOH) present in 1.50 * 10^24 molecules of NaOH.