whatmass of (NH4)3SO4 is required to prepare 500cm3 of m/4 solution
what is m/4 solution?
Does m stand for molality of molarity? You can't get away forever with not using caps. m stands for molality. M stands for molarity. Since you don't use caps to start a sentence we don't know if this is 1/4 m or 1/4 M.
To calculate the mass of (NH4)3SO4 required to prepare a 500 cm³ m/4 solution, we need to follow these steps:
Step 1: Determine the molarity of the desired solution.
The molarity (M) is defined as the number of moles of solute per liter of solution. In this case, we want a molarity of 1/4, which means 1/4 mole per liter or 0.25 moles per liter.
Step 2: Convert the desired solution volume to liters.
Given that the volume is 500 cm³, we need to convert it to liters by dividing by 1000. Therefore, the volume in liters is 0.5 liters.
Step 3: Calculate the number of moles of (NH4)3SO4 needed.
Using the equation M = moles/volume, you can rearrange the equation to solve for moles: moles = M × volume.
moles = 0.25 mol/L × 0.5 L
moles = 0.125 moles
Step 4: Determine the molar mass of (NH4)3SO4.
(NH4)3SO4 consists of one nitrogen atom (N) with a molar mass of 14.01 g/mol, three hydrogen atoms (H) with a molar mass of 1.01 g/mol each, four oxygen atoms (O) with a molar mass of 16.00 g/mol each, and twelve hydrogen atoms (H) with a molar mass of 1.01 g/mol each.
Molar mass of (NH4)3SO4 = (3 × 14.01 g/mol) + (12 × 1.01 g/mol) + (4 × 16.00 g/mol)
Molar mass of (NH4)3SO4 = 132.14 g/mol
Step 5: Calculate the mass of (NH4)3SO4 needed.
To calculate the mass, we use the equation mass = moles × molar mass.
mass = 0.125 moles × 132.14 g/mol
mass ≈ 16.52 g
Therefore, approximately 16.52 grams of (NH4)3SO4 is required to prepare 500 cm³ of m/4 solution.