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2 years ago

Calculate moles of Fe in 5.22 X 10^22 atoms of Fe?

Chemistry

1 answer:

Setler [38]2 years ago

5 0

Answer:

The amount of moles of Fe in 5.22*10²² atoms of Fe is 0.0867.

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension, so it is considered a pure number that allows describing a physical characteristic without an explicit dimension or unit of expression. Avogadro's number applies to any substance.

Then you can apply the following rule of three: if 6.023*10²³ atoms are contained in 1 mole, 5.22*10²² atoms are contained in how many moles?

$amount of moles=\frac{5.22*10^{22}atoms * 1 mole}{6.023*10^{23}atoms}$

amount of moles= 0.0867

The amount of moles of Fe in 5.22*10²² atoms of Fe is 0.0867.

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A student has 70.5 mL of a 0.463 M aqueous solution of sodium bromide. The density of the solution is 1.22 g/mL. Find the follow

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a.) 86.01 g.

b.) 3.36 g.

c.) 0.394 m ≅ 0.40 m.

d.) 4.77%.

e.) 3.9%.

Explanation:

a.) mass of the solution:

The density of the solution is the mass per unit volume.

∵ Density of solution = (mass of solution)/(volume of the solution).

∴ Mass of the solution = (density of solution)*(volume of the solution) = (1.22 g/mL)*(70.5 mL) = 86.01 g.

b.) grams of sodium bromide :

Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

∵ M = (no. of moles of NaBr)/(Volume of the solution (L))

∴ no. of moles of NaBr = M*(Volume of the solution (L)) = (0.463 M )*(0.0705 L) = 0.0326 mol.

∵ no. of moles of NaBr = (mass of NaBr)/(molar mass of NaBr)

∴ mass of NaBr = (no. of moles of NaBr)*(molar mass of NaBr) = (0.0326 mol)*(102.894 g/mol) = 3.36 g.

c.) molality of the solution:

Molality (m) is defined as the no. of moles of solute dissolved per 1.0 kg of the solvent.

∵ m = (no. of moles of NaBr)/(mass of the soluvent (kg))

no. of moles of NaBr = 0.0326 mol,

mass of solvent = mass of the solution - mass of NaBr = 86.01 g - 3.36 g = 82.65 g = 0.08265 kg.

∴ m = (no. of moles of NaBr)/(mass of the soluvent (kg)) = (0.0326 mol)/(0.08265 kg) = 0.394 m ≅ 0.40 m.

d.) % (m/v) of the solution:

∵ (m/v)% = [(mass of solute) /(volume of the solution)]* 100

∴ (m/v)% = [(3.36 g)/(70.5 mL)]* 100 = 4.77%.

e.) % (m/m) of the solution:

∵ (m/m)% = [(mass of solute) /(mass of the solution)]* 100

∴ (m/m)% = [(3.36 g)/(86.01 g)] * 100 = 3.9 %.

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