Question

*What are the respective concentrations (M) of Fe3+ and I- afforded by dissolving 0.300 mol FeI3 in water and diluting it to 750 mL?
* Which solution contains the largest number of moles of chloride ions?
A) 10.0 mL of 0.400 M BaCl2
B) 2.00 mL of 1.000 M NaCl C) 7.50 mL of 0.200 M FeCl3
D) 15.00 mL of 0.400 M KCl E) 20.00 mL of 0.100 M CaCl2
13. A 51.0 mL aliquot of HCl(aq) of unknown concentration was titrated with 0.226 M NaOH(aq). It took 102.4 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was...?

Answer 1

Answer on Question #55378 – Chemistry – General chemistry

Question:

1) What are the respective concentrations (M) of Fe³⁺ and I⁻ afforded by dissolving 0.300 mol FeI₃ in water and diluting it to 750 mL?

2) Which solution contains the largest number of moles of chloride ions?

A) 10.0 mL of 0.400 M BaCl₂

B) 2.00 mL of 1.000 M NaCl

C) 7.50 mL of 0.200 M FeCl₃

D) 15.00 mL of 0.400 M KCl

E) 20.00 mL of 0.100 M CaCl₂

3) A 51.0 mL aliquot of HCl(aq) of unknown concentration was titrated with 0.226 M NaOH(aq). It took 102.4 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was...?

1) FeI₃ = Fe³⁺ + 3I⁻

$$c = \frac {n}{V}$$

Concentration or molarity is most expressed by lowercase c, 1 molar = 1 M = 1 mol/L. n is the amount of the solute in moles in the volume V (in litres)

$$V = 750 \mathrm{~mL} = 0.75 \mathrm{~L}$$ $$\mathrm{n} \left(\mathrm{Fe}^{3+}\right) = \mathrm{n} \left(\mathrm{FeI}_3\right) = 0.300 \mathrm{~mol}$$ $$\mathrm{c} \left(\mathrm{Fe}^{3+}\right) = \mathrm{n} \left(\mathrm{Fe}^{3+}\right) / \mathrm{V} = 0.300 / 0.75 = 0.4 \mathrm{~mol/L}$$ $$\mathrm{n} \left(\mathrm{I}^{-}\right) = 3 \times \mathrm{n} \left(\mathrm{FeI}_3\right) = 3 \times 0.300 \mathrm{~mol} = 0.900 \mathrm{~mol}$$ $$\mathrm{c} \left(\mathrm{I}^{-}\right) = \mathrm{n} \left(\mathrm{I}^{-}\right) / \mathrm{V} = 0.900 / 0.75 = 1.2 \mathrm{~mol/L}$$

The Answer is c (Fe³⁺) = 0.4 M, c (I⁻) = 1.2 M

2)

3) NaOH + HCl = NaCl + H₂O

$$\mathrm{n} (\mathrm{HCl}) = \mathrm{n} (\mathrm{NaOH}) = \mathrm{c} (\mathrm{NaOH}) \times \mathrm{V} (\mathrm{NaOH}) = 0.226 \times 0.1024 = 0.023 \mathrm{~mol}$$ $$\mathrm{c} (\mathrm{HCl}) = \mathrm{n} (\mathrm{HCl}) / \mathrm{V} (\mathrm{HCl}) = 0.023 / 0.051 = 0.453 \mathrm{~mol/L}$$

The Answer is 0.453 M

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