Q4.19P How many moles and numbers of io... [FREE SOLUTION]

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Chapter 4: Q4.19P (page 173)

How many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) 88 mL of 1.75 M magnesium chloride
(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L
(c) 1.65 L of a solution containing 8.83 x 10²¹ formula units of cesium nitrate 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) 0.154 mol Mg²⁺, 0.308 mol Cl^-

9.27 x 10²² Mg²⁺ ions, 1.85 x 10²³ Cl^- ions

(b) 4.13 x 10^-4mol Al³⁺, 6.19 x 10^-4mol SO₄^2-

2.49 x 10²⁰ Al³⁺ ions, 3.73 x 10²⁰ SO₄^2-ions

1.46 x 10²²Cs⁺ ions, 1.46 x 10²²NO₃^-ions

Step by step solution

the given values are:

(a) 88 mL of 1.75 M magnesium chloride

(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L

Write down the balanced dissociation reaction and find out the moles and no. of ions as follows:

(a) Dissociation reaction of magnesium chloride is:

$M g C l_{2} (s) 鈫� M g^{2 +} (a q) + 2 C l^{-} (a q)$

${MolesofMg}^{2 +} = (88 m l) x (\frac{1 L}{1000 m l}) x (\frac{1.75 m o l M g C l_{2}}{L i t}) (\frac{1 m o l M g^{2 +}}{1 m o l M g C l_{2}}) = 0.154 m o l M g^{2 +} m o l e s o f C l^{-} = (88 m l) x (\frac{1 L}{1000 m l}) x (\frac{1.75 m o l M g C l_{2}}{L i t}) (\frac{2 m o l C l^{-}}{1 m o l M g C l_{2}}) = 0.308 m o l C l^{-} M g^{2 +} i o n s = (0.154 m o l M g^{2 +}) x (\frac{6.022 x 10^{23} M g^{2 +} i o n}{1 m o l M g^{2 +}}) = 9.27 x 10^{22} M g^{2 +} i o n s C l^{-} i o n s = (0.308 m o l C l^{-}) x (\frac{6.022 x 10^{23} C l^{-} i o n}{1 m o l C l^{-}}$

(b) Dissociation reaction ofrole="math" localid="1663304705230" $A l_{2} ({SO}_{4})_{3} (s)$ is:

$A l_{2} (S O_{4})_{3} (s) 鈫� 2 A l^{3 +} (a q) + 3 S {O_{4}}^{2 -} (a q)$

$m o l e s o f A l^{3 +} = (321 m l) (\frac{1 L}{1000 m l}) (\frac{. 22 g m A l_{2} {(S O_{4})}_{3}}{L i t}) (\frac{1 m o l A l_{2} {(S O_{4})}_{3}}{342.17 g A l_{2} {(S O_{4})}_{3}}) (\frac{2 m o l A l^{3 +}}{1 m o l A l_{2} {(S O_{4})}_{3}}) = 4.13 x 10^{- 4} m o l A l^{3 +} m o l e s o f S {O_{4}}^{2 -} = (321 m l) (\frac{1 L}{1000 m l}) (\frac{. 22 g m A l_{2} {(S O_{4})}_{3}}{L i t}) (\frac{1 m o l A l_{2} {(S O_{4})}_{3}}{342.17 g A l_{2} {(S O_{4})}_{3}}) (\frac{3 m o l S {O_{4}}^{2 -}}{1 m o l A l_{2} {(S O_{4})}_{3}}) = 6.19 x 10^{- 4} m o l S {O_{4}}^{2 -} A l^{3 +} i o n s = (4.13 x 10^{- 4} m o l A l^{3 +}) x (\frac{6.022 x 10^{23} A l^{3 +} i o n}{1 m o l A l^{3 +}}) = 2.49 x 10^{20} A l^{3 +} i o n s S {O_{4}}^{2 -} i o n s = (6.19 x 10^{- 4} m o l S {O_{4}}^{2 -}) x (\frac{6.022 x 10^{23} S {O_{4}}^{2 -} i o n}{1 m o l S {O_{4}}^{2 -}}) = 3.73 x 10^{20} S {O_{4}}^{2 -} i o n s \$

$C s N O_{3} (s) 鈫� C s^{+} (a q) + N {O_{3}}^{-} (a q)$

$C s^{+} i o n s = (1.65 l i t) (\frac{8.83 x 10^{21} F . U . C s N O_{3}}{L i t}) (\frac{1 C s^{+} i o n}{1 F . U . C s N O_{3}}) = 1.46 x 10^{22} C s^{+} i o n s N {O_{3}}^{-} i o n s = (165 m l) (\frac{1 L}{1000 m l}) (\frac{8.83 x 10^{21} F . U . C s N O_{3}}{L i t}) (\frac{1 N {O_{3}}^{-} i o n}{1 F . U . C s N O_{3}}) l = 1.46 x 10^{22} N {O_{3}}^{-} i o n s \ begingathered C s^{+} m o l e s = (9.02 x 10^{21} C s^{+} i o n) x (\frac{1 m o l C s^{+}}{6.022 x 10^{23} C s^{+} i o n}) = 2.42 x 10^{- 2} m o l C s^{+} N {O_{3}}^{-} m o l e s = (9.02 x 10^{21} N {O_{3}}^{-} i o n) x (\frac{1 m o l N {O_{3}}^{-}}{6.022 x 10^{23} N {O_{3}}^{-} i o n}) = 2.42 x 10^{- 2} m o l N {O_{3}}^{-}$

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91影视

How many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) 88 mL of 1.75 M magnesium chloride
(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L
(c) 1.65 L of a solution containing 8.83 x 10²¹ formula units of cesium nitrate per liter.

Short Answer

Step by step solution

the given values are:

Write down the balanced dissociation reaction and find out the moles and no. of ions as follows:

One App. One Place for Learning.

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Recommended explanations on Chemistry Textbooks

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91影视

How many moles and numbers of ions of each type are present in the following aqueous solutions?(a) 88 mL of 1.75 M magnesium chloride(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L(c) 1.65 L of a solution containing 8.83 x 1021 formula units of cesium nitrate per liter.

Short Answer

Step by step solution

the given values are:

Write down the balanced dissociation reaction and find out the moles and no. of ions as follows:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Physical Chemistry

The Earths Atmosphere

Making Measurements

Chemical Reactions

Kinetics

Study anywhere. Anytime. Across all devices.

How many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) 88 mL of 1.75 M magnesium chloride
(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L
(c) 1.65 L of a solution containing 8.83 x 10²¹ formula units of cesium nitrate per liter.