Calculate the volume of solution of 3.6 M NH4CI needed to prepare 350.0ml of 1.24 M NH4CI.

How many grams of ammonium chloride are in the solution you created in the above problem.

To calculate the volume of the 3.6 M NH4CI solution needed to prepare 350.0 ml of 1.24 M NH4CI, we can use the formula:

C₁V₁ = C₂V₂

Where:
C₁ = concentration of the stock solution
V₁ = volume of the stock solution
C₂ = concentration of the desired solution
V₂ = volume of the desired solution

Let's plug in the values:

C₁ = 3.6 M
V₁ = ?
C₂ = 1.24 M
V₂ = 350.0 ml

3.6 M x V₁ = 1.24 M x 350.0 ml

Now, let's solve for V₁:

V₁ = (1.24 M x 350.0 ml) / 3.6 M
V₁ = 120.56 ml (rounded to two decimal places)

Therefore, you would need approximately 120.56 ml of the 3.6 M NH4CI solution to prepare 350.0 ml of 1.24 M NH4CI.

Now let's calculate the mass of ammonium chloride in the solution you created.

The molar mass of NH4CI (ammonium chloride) is:
NH4CI = (1 x 14.01 g/mol) + (4 x 1.00784 g/mol) + (1 x 35.453 g/mol)
NH4CI = 53.49 g/mol

To calculate the mass of ammonium chloride, we can use the formula:

Mass = concentration x volume x molar mass

Let's plug in the values:

Concentration = 1.24 M
Volume = 350.0 ml
Molar Mass of NH4CI = 53.49 g/mol

Mass = 1.24 M x 350.0 ml x 53.49 g/mol
Mass = 23,388.6 g (rounded to one decimal place)

Therefore, the amount of ammonium chloride in the solution you created is approximately 23,388.6 grams.

To calculate the volume of solution needed, we can use the following equation:

(M1)(V1) = (M2)(V2)

Where:
M1 = initial molarity (3.6 M)
V1 = initial volume (unknown)
M2 = final molarity (1.24 M)
V2 = final volume (350.0 ml)

Rearranging the equation to solve for V1:

V1 = (M2)(V2) / M1

Plugging in the given values:

V1 = (1.24 M)(350.0 ml) / 3.6 M
V1 = 120.56 ml

Therefore, you would need 120.56 ml of the 3.6 M NH4CI solution to prepare 350.0 ml of 1.24 M NH4CI.

To calculate the grams of ammonium chloride in the solution, we can use the formula:

mass = molarity * volume * molar mass

The molar mass of NH4CI (ammonium chloride) is 53.49 g/mol.

mass = 1.24 M * 350.0 ml * 53.49 g/mol
mass = 22920.6 g

Therefore, there are 22920.6 grams of ammonium chloride in the solution you created in the above problem.

2. mL1 x M1 = ml2 x M2

mL1 x 3.6 = 350 x 3.6

3. mols NH4Cl = M x L
grams = mols x molar mass.