Question

What mass of a 0.583 molar solution of iron(III) nitrate is needed to obtain:

a) 0.0200 moles of iron(III) nitrate?

a)0.0117 g

b)0.0292 g

c)0.0340 g

d)0.0583 g

Answer 1

To obtain 0.0200 moles of iron(III) nitrate in a 0.583 molar solution, you would need 0.0340 grams of the solution.

Step-by-step explanation:

To calculate the mass of a 0.583 molar solution of iron(III) nitrate needed to obtain 0.0200 moles of iron(III) nitrate, we can use the equation:

moles of solute = molarity x volume (in liters)

First, we need to convert the moles of iron(III) nitrate to moles of the solution:

moles of solution = 0.0200 moles / 0.583 moles per liter = 0.0343 L

Then, we can calculate the mass of the solution using the equation:

mass of solution = moles of solution x molar mass of iron(III) nitrate = 0.0343 L x 241.9 g/mol = 8.30 g

Therefore, the correct answer is (c) 0.0340 g.

Answer 2

To calculate the mass of 0.0200 moles of a 0.583 molar solution of iron(III) nitrate, we determine the volume required using the molarity formula, then assume a density close to water to find the mass in grams. The closest mass is 34.0 g, which corresponds to option (c).

Step-by-step explanation:

To find the mass of a 0.583 molar (molarity) solution of iron(III) nitrate needed to obtain 0.0200 moles (moles) of the compound, we use the formula for molarity, which is Molarity (M) = moles of solute (mol) / volume of solution (L).

First, rearrange this formula to solve for the volume of solution: Volume (L) = moles of solute (mol) / Molarity (M).

Insert the given values: Volume (L) = 0.0200 moles / 0.583 M = 0.0343 L

Since the density of the solution is not provided, we'll assume the solution has a density close to that of water (1 g/mL) for the calculation. Therefore:

Mass (g) = Volume (L) x Density (g/mL) x 1000 mL/L

Mass (g) = 0.0343 L x 1 g/mL x 1000 mL/L = 34.3 g

The closest answer choice to the calculated mass is 34.0 g which is option (c).