Q4.18P Question:聽How many moles and nu... [FREE SOLUTION]

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Chapter 4: Q4.18P (page 170)

Question:How many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) 130 mL of 0.45 M aluminum chloride
(b) 9.80 mL of a solution containing 2.59 g lithium sulfate/L
(c) 245 mL of a solution containingformula units of potassium bromide per liter.

Short Answer

Expert verified

Answer

The moles and numbers of ions of each type are present in the following aqueous solutions are:

(a) 0.059 mol AI³⁺ 0.176 mol CI^-

$355 脳 10^{22} A I^{3 +} i o n s, 1.06 脳 10^{23} C I^{-}^{}$

$(b) 4.62 脳 10^{- 4} m o l L i^{+}, 2.31 脳 10^{- 4} m o l s {o_{4}}^{2 -} 2.78 脳 10^{20} m o l L i^{+} i o n s, 1.39 脳 10^{20} m o l s {o_{4}}^{2 -} i o n s (c) 1.50 脳 10^{- 2} m o l K^{+}, 1.50 脳 10^{- 2} m o l {B_{r}}^{-}$

Step by step solution

Calculate moles of ions released when dissolve in water

The dissociation reaction of aluminium chlorideis:

$A I C I_{3} (s) 鈫� A I^{3 +} (aq) + 3 C I^{-} (aq)$

The moles of $A I^{3 +}$ is calculated as:

$m o l e s o f A I^{3 +} = (130 mI) (\frac{1 L}{1000 ml}) (\frac{0.45 moI {AICI}_{3}}{Lit}) (\frac{1 {molH}^{+}}{1 mol {AICI}_{3}}) = 0.059 m o l A I^{3 +}$

The moles of $C I^{-}$ is calculated as:

localid="1663932460834" $m o l e s o f A I^{3 +} = (130 mI) (\frac{1 L}{1000 ml}) (\frac{0.45 moI {AICI}_{3}}{Lit}) (\frac{3 m o l C I^{-}}{1 mol {AICI}_{3}}) = 0.1769 m o l C I^{-}$

Calculate number of ions of AI3+ and CI-

The number of ions of AI³⁺ is:

$A I^{3 +} i o n s = (0.059 m o l A I^{3 +}) (\frac{6.022 脳 10^{23} A I^{3 +} i o n .}{1 m o l A I^{3 +}} = 3.55 脳 10^{22} m o l A I^{3 +} i o n s$

The number of ions of CI^-is:

role="math" localid="1663310371661" $m o l e s o f C I^{-} = (0.1769 m o l C I^{-}) (\frac{6.022 脳 10^{23} A I^{3 +} i o n .}{1 m o l C I^{-}} = 1.06 脳 10^{23} m o l C I^{-} i o n s$

Calculate moles of ions released when dissolve in water

The dissociation reaction of lithium sulfateis:

$L i_{2} S O_{4} (s) 鈫� 2 L I^{+} (aq) + S {O_{4}}^{2 -} (aq)$

The moles of $L i^{+}$ is calculated as:

$m o l e s o f L i^{+} = (9.80 mI) (\frac{1 L}{1000 ml}) (\frac{2.59 g {Li}^{2} S {O_{4}}^{}}{Lit}) (\frac{1 mole of {Li}^{+}}{109.95 g {Li}^{2} S {O_{4}}^{}}) = 4.62 脳 10^{- 4} L i^{+} i o n s$

The moles of $S {O_{4}}^{2 -}$ is calculated as:

$m o l e s o f S {O_{4}}^{2 -} i o n s = (9.80 m I) (\frac{1 L}{1000 m l}) (\frac{2.59 g {Li}^{2} S {O_{4}}^{}}{Lit}) (\frac{1 m o l e o f {Li}^{2} {SO}_{4}}{109.95 g {Li}^{2} S {O_{4}}^{}}) = 2.31 脳 10^{- 4} S {O_{4}}^{2 -} i o n s$

Calculate number of ions of SO42- and Li+

The number of Li⁺ ions of is:

$L i^{+} i o n s = (4.62 脳 10^{- 4} m o l L i^{+}) (\frac{6022 脳 10^{23} C I^{-} i o n s .}{1 m o l C I^{-} i o n}) 2.78 脳 10^{20} L i^{+} i o n s$

The number of $S {O_{4}}^{2 -}$ ions of is:

role="math" localid="1663309282903" $S {O_{4}}^{2 -} i o n s = (2.31 脳 10^{- 4} S {O_{4}}^{2 -} ions) (\frac{6022 脳 10^{23} S {O_{4}}^{2 -} i o n s .}{1 m o l S {O_{4}}^{2 -} i o n}) = 1.39 脳 10^{20} S {O_{4}}^{2 -} i o n s$

Calculate moles of ions released when dissolve in water

The dissociation reaction of potassium bromideis:

$K B r (s) 鈫� K^{+} (aq) + B r^{-} (aq)$

The moles of K⁺is calculated as:

$k^{+} i o n s = (9.02 脳 10^{21} k^{+} ions) (\frac{1 m o l k^{+}}{6022 脳 10^{23} k^{+} i o n s .}) = 1.50 脳 10^{23} 脳 10^{2} m o l k^{+}$

The moles of Br ^-is calculated as:

role="math" localid="1663308469248" $B r i o n s = (9.02 脳 10^{21} {Br}^{-} ions) (\frac{1 m o l b r^{-}}{6022 脳 10^{23} B r i o n s .}) = 1.50 脳 10^{23} 脳 10^{2} m o l B r^{-}$

Calculate number of ions of and

The number of K⁺ ions of is:

$K^{+} i o n s = (245 mI) (\frac{1 L}{1000 ml}) (\frac{3.68 脳 10^{22} F . U . KBr}{Lit}) (\frac{1 B r i o n s}{1 F . U . KBr}) = 9.02 脳 10^{21} K^{+} i o n s$

The number of B_r^-ions of is:

$B r i o n s = (245 mI) (\frac{1 L}{1000 ml}) (\frac{3.68 脳 10^{22} F . U . KBr}{Lit}) (\frac{1 B r i o n s}{1 F . U . KBr}) = 9.02 脳 10^{21} B r^{-} i o n s$

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91影视

Question:How many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) 130 mL of 0.45 M aluminum chloride
(b) 9.80 mL of a solution containing 2.59 g lithium sulfate/L
(c) 245 mL of a solution containingformula units of potassium bromide per liter.

Short Answer

Step by step solution

Calculate moles of ions released when dissolve in water

Calculate number of ions of AI3+ and CI-

Calculate moles of ions released when dissolve in water

Calculate number of ions of SO42- and Li+

Calculate moles of ions released when dissolve in water

Calculate number of ions of and

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Most popular questions from this chapter

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91影视

Question:How many moles and numbers of ions of each type are present in the following aqueous solutions?(a) 130 mL of 0.45 M aluminum chloride(b) 9.80 mL of a solution containing 2.59 g lithium sulfate/L(c) 245 mL of a solution containingformula units of potassium bromide per liter.

Short Answer

Step by step solution

Calculate moles of ions released when dissolve in water

Calculate number of ions of AI3+ and CI-

Calculate moles of ions released when dissolve in water

Calculate number of ions of SO42- and Li+

Calculate moles of ions released when dissolve in water

Calculate number of ions of and

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Chemical Reactions

The Earths Atmosphere

Chemical Analysis

Physical Chemistry

Nuclear Chemistry

Study anywhere. Anytime. Across all devices.

Question:How many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) 130 mL of 0.45 M aluminum chloride
(b) 9.80 mL of a solution containing 2.59 g lithium sulfate/L
(c) 245 mL of a solution containingformula units of potassium bromide per liter.