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3 years ago

What are the respective concentrations (m) of fe3+ and i- afforded by dissolving 0.200 mol fei3 in water and diluting to 725 ml?

Chemistry

1 answer:

Natalka [10]3 years ago

7 0

Molarity = moles / liter of solution

Given, Moles of Fe³⁺ = 0.200

Volume of solution = 725 ml = 0.725 L

Conversion factor: 1000 ml = 1L

Molarity = 0.200 / 0.725 L = 0.275 M

The dissociation of Fel₃ in water is ad follows:

Fel₃ → Fe³⁺ + 3l⁻

1 mole of Fel₃ gives 1 mole of Fe³⁺ ions and 3 moles of l⁻

Since the solution is 0.275 M of Fel₃, so there are 0.275 M of Fe³⁺ ions,

and (3 x 0.275 M) = 0.825 M of l⁻ ions.

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3 years ago

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3 years ago

Read 2 more answers

If 5 grams of hydrogen reacted with 40 games of oxygen to form water, how much of water would be formed?

GREYUIT [131]

2H₂ + O₂ = 2H₂O

n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol

n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol

H₂ : O₂ = 2 : 1

2.5 : 1.25 = 2 : 1

n(H₂O)=n(H₂)=2n(O₂)=2.5 mol

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m(H₂O)=2.5mol*18.0g/mol=45.0 g

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3 years ago

20 ml of an approximately 10% aqueous solution ofethylamine, ch3ch2nh2, is titrated with 0.3000 maqueous hcl.which indicator wou

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3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

3 years ago