If I want to prepare 5.0 liters of a 0.25 M solution of NH4Cl, how many grams of the salt should I weight out?

66.86g

Well, let's calculate that, shall we?

First, we need to find the molar mass of NH₄Cl. Nitrogen (N) has a molar mass of approximately 14.01 g/mol, hydrogen (H) has a molar mass of approximately 1.01 g/mol, and chlorine (Cl) has a molar mass of approximately 35.45 g/mol. Adding these up, we get:

(14.01 g/mol * 1) + (1.01 g/mol * 4) + (35.45 g/mol * 1) = 53.49 g/mol

So, now we know that NH₄Cl has a molar mass of approximately 53.49 g/mol.

Now, we can use the formula for molarity to find the number of moles of NH₄Cl required:

Molarity (M) = moles of solute / liters of solution

Rearranging the formula, we get:

moles of solute = Molarity (M) * liters of solution

Plugging in the values, we have:

0.25 M * 5.0 L = 1.25 moles of NH₄Cl

Now, we can finally calculate the mass of NH₄Cl by multiplying the number of moles by the molar mass:

1.25 moles * 53.49 g/mol = 66.86 grams

So, you would need to weigh out approximately 66.86 grams of NH₄Cl to prepare your 0.25 M solution in 5.0 liters. Happy weighing!

To determine the number of grams of NH4Cl you should weigh out, you need to use the formula:

Mass (in grams) = molar concentration (in mol/L) × volume (in L) × molar mass (in g/mol)

Given:
- Molar concentration of NH4Cl: 0.25 M
- Volume of solution: 5.0 L

Now let's calculate the mass of NH4Cl:

1. Find the molar mass of NH4Cl:
- Molar mass of N = 14.01 g/mol
- Molar mass of H = 1.01 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of NH4Cl = (1 × 14.01) + (4 × 1.01) + 35.45 = 53.49 g/mol

2. Substitute the values into the formula:
Mass (in grams) = 0.25 mol/L × 5.0 L × 53.49 g/mol

3. Solve for the mass of NH4Cl:
Mass (in grams) = 0.25 × 5.0 × 53.49 ≈ 67.4 grams

Therefore, you should weigh out approximately 67.4 grams of NH4Cl to prepare 5.0 liters of a 0.25 M solution.

To find out how many grams of NH4Cl you need to weigh out, you need to use the formula:

moles = concentration × volume

First, let's convert the volume from liters to milliliters since the concentration is given in moles per liter. There are 1000 milliliters in a liter, so 5.0 liters is equal to 5000 milliliters.

Now, let's substitute the given values into the formula:

moles = 0.25 mol/L × 5000 mL

Next, we need to convert milliliters to liters:

moles = 0.25 mol/L × 5.0 L

Now, we can solve for the number of moles:

moles = 1.25 moles

Lastly, we need to determine the molar mass of NH4Cl. The molar mass of NH4 is 14.01 g/mol and the molar mass of Cl is 35.45 g/mol. Adding these two together gives a molar mass of 14.01 + 35.45 = 49.46 g/mol.

Now, we can calculate the number of grams of NH4Cl needed:

grams = moles × molar mass

grams = 1.25 moles × 49.46 g/mol

Therefore, you would need to weigh out approximately 61.83 grams of NH4Cl to prepare a 0.25 M solution in 5.0 liters.

How many mols do you want? That's M x L = ?

Then mols = grams/molar mass
You know mols and molar mass, solve for grams.