Question

What is the concentration of iron (III) nitrate in a solution prepared by dissolving 44.1 g of Fe(NO3)3

in 478 mL of water

A- 0.000381 M Fe(NO3)3
B- 92.3 M Fe(NO3)3
C- 2.62 M Fe(NO3)3
D- 0.381 M Fe(NO3)3
E- 0.182 M Fe(NO3)3

Answer 1

To find the concentration of iron (III) nitrate, calculate the moles by dividing the mass by the molar mass, then divide the result by the volume in liters. The concentration of iron (III) nitrate is 0.381 M after dissolving 44.1 g in 478 mL of water, making option D the correct answer.

Step-by-step explanation:

The question asks for the concentration of iron (III) nitrate in a solution after dissolving 44.1 g of Fe(NO3)3 in 478 mL of water. First, we need to calculate the number of moles of Fe(NO3)3 dissolved. The molar mass of Fe(NO3)3 (the sum of one iron atom, three nitrogen atoms, and nine oxygen atoms) is approximately 241.86 g/mol. The number of moles of Fe(NO3)3 is given by the formula:

moles of Fe(NO3)3 = mass (g) ÷ molar mass (g/mol)

moles of Fe(NO3)3 = 44.1 g ÷ 241.86 g/mol = 0.182 mol

Now to find the concentration in molarity (M), we divide the moles by the volume of the solution in liters:

concentration (M) = moles of solute (mol) ÷ volume of solution (L)

Since 478 mL is equal to 0.478 L, the concentration is:

concentration (M) = 0.182 mol ÷ 0.478 L = 0.381 M

Therefore, the correct answer is: D- 0.381 M Fe(NO3)3

Answer 2

0.381 mol/L

Step-by-step explanation:

"Concentration" is another word for Molarity (at least in this context). Molarity (M) is defined as:

M = mol/L

To find the concentration/molarity, you'll need to find the number of moles of iron (III) nitrate divided by the liters of the solution.

1. Convert grams to moles

You first need to calculate the molar mass of iron (III) nitrate, so you can convert the grams to moles. To do this, you add the masses of each atom in the compound (the masses of each atom can be found from the periodic table).

Fe (iron) has a mass of 55.845 g

N (nitrogen) has a mass of 14.007 g

O (oxygen) has a mass of 16 g

(1 x 55.845) + 3(14.007 + (16 x 3)) = 241.88 g/mol

193.9 is iron (III) nitrate's molar mass, so now you can divide the grams you have by its molar mass to get the number of moles.

44.1/241.88 = 0.1823 mol

2. Convert mL to L

There are 1000 milliliters in 1 liter, so you can divide the number of mL by 1000 to get the number of liters.

478/1000 = 0.478 L

3. Calculate Molarity/concentration

M = mol/L

M = 0.1823/0.478

M = 0.381