OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781285460420
Author: John W. Moore; Conrad L. Stanitski
Publisher: Cengage Learning US
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Chapter 13, Problem 47QRT

(a)

Interpretation Introduction

Interpretation:

The molarity of the solute 6.18gMgNH4PO4 in 250mL solution has to be calculated.

Concept introduction:

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

  Molarity=MolesofsoluteVolume of solution in L

The number of moles of any substance can be determined using the equation.

  Numberofmoles=GivenmassofthesubstanceMolarmass

Moles: One mole is equivalent to the mass of the substance consists of same number of units equal to the atoms present in 12g of 12C.

(a)

Expert Solution
Check Mark

Answer to Problem 47QRT

The molarity obtained as 0.18 M.

Explanation of Solution

Given mass is 6.18g; Volume is 250 mL; Molar mass MgNH4PO4 is137.31 g/mol.

Calculate the number of moles:

  Number of moles = MassMolarmass=6.18g137.31 g/mol=45.00×10-3moles.

Calculate the molarity:

   Molarity =Number of molesvolume=45.00×10-3moles250ml=0.18 M.

Therefore, the molarity of the solute 6.18gMgNH4PO4 in 250mL solution was 0.18 M.

(b)

Interpretation Introduction

Interpretation:

The molarity of the solute 16.8gNaCH3COO in 300mL solution has to be calculated.

Concept introduction:

Refer part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 47QRT

The molarity obtained as 0.6826M.

Explanation of Solution

Given mass is 16.8g; Volume is 300 mL; Molar mass NaCH3COOis 82.0326 g/mol.

Calculate the number of moles:

  Number of moles = MassMolarmass=16.8g82.0326 g/mol=204.79×10-3moles.

Calculate the molarity:

   Molarity =Number of molesvolume=204.79×10-3moles300ml=0.6826M.

Therefore, the molarity of the solute 16.8gNaCH3COO in 300mL solution was 0.6826M.

(c)

Interpretation Introduction

Interpretation:

The molarity of the solute 2.50gCaC2O4 in 750mL solution has to be calculated.

Concept introduction:

Refer part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 47QRT

The molarity obtained as 0.026M.

Explanation of Solution

Given mass is 2.50g; Volume is 750 mL; Molar mass CaC2O4is 128.097 g/mol.

Calculate the number of moles:

  Number of moles = MassMolarmass=2.50g128.097 g/mol=19.5×10-3moles.

Calculate the molarity:

   Molarity =Number of molesvolume=19.5×10-3moles.750ml=0.026M.

Therefore, the molarity of the solute 16.8gCaC2O4 in 300mL solution was 0.026M.

(d)

Interpretation Introduction

Interpretation:

The molarity of the solute 2.20g(NH4)2SO4 in 400mL solution has to be calculated.

Concept introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 47QRT

The molarity obtained as 0.0415M.

Explanation of Solution

Given mass is 2.20g; Volume is 400 mL; Molar mass (NH4)2SO4is132.14 g/mol.

Calculate the number of moles:

  Number of moles = MassMolarmass=2.20g132.14 g/mol=16.6×10-3moles.

Calculate the molarity:

   Molarity =Number of molesvolume=16.6×10-3moles400ml=0.0415M.

Therefore, the molarity of the solute 2.20g(NH4)2SO4 in 400mL solution was 0.0415M.

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Chapter 13 Solutions

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

Ch. 13.6 - Prob. 13.4PSPCh. 13.6 - Prob. 13.8ECh. 13.6 - Drinking water may contain small quantities of...Ch. 13.6 - Prob. 13.9CECh. 13.6 - A 500-mL bottle of Evian bottled water contains 12...Ch. 13.6 - The mass fraction of gold in seawater is 1 103...Ch. 13.6 - Prob. 13.6PSPCh. 13.6 - Prob. 13.7PSPCh. 13.6 - Prob. 13.8PSPCh. 13.6 - Prob. 13.9PSPCh. 13.6 - Prob. 13.12ECh. 13.6 - Prob. 13.13CECh. 13.7 - The vapor pressure of an aqueous solution of urea....Ch. 13.7 - Prob. 13.14ECh. 13.7 - Prob. 13.15ECh. 13.7 - Prob. 13.11PSPCh. 13.7 - Suppose that you are closing a cabin in the north...Ch. 13.7 - A student determines the freezing point to be 5.15...Ch. 13.7 - Prob. 13.17CECh. 13.7 - Prob. 13.13PSPCh. 13.9 - Prob. 13.18CECh. 13.10 - Prob. 13.19ECh. 13.10 - Prob. 13.20ECh. 13 - Prob. 1QRTCh. 13 - Prob. 2QRTCh. 13 - Prob. 3QRTCh. 13 - Prob. 4QRTCh. 13 - Prob. 5QRTCh. 13 - Prob. 6QRTCh. 13 - Prob. 7QRTCh. 13 - Prob. 8QRTCh. 13 - Prob. 9QRTCh. 13 - Prob. 10QRTCh. 13 - Prob. 11QRTCh. 13 - Prob. 12QRTCh. 13 - Prob. 13QRTCh. 13 - Prob. 14QRTCh. 13 - Beakers (a), (b), and (c) are representations of...Ch. 13 - Prob. 16QRTCh. 13 - Simple acids such as formic acid, HCOOH, and...Ch. 13 - Prob. 18QRTCh. 13 - Prob. 19QRTCh. 13 - Prob. 20QRTCh. 13 - Prob. 21QRTCh. 13 - Prob. 22QRTCh. 13 - Prob. 23QRTCh. 13 - Prob. 24QRTCh. 13 - Prob. 25QRTCh. 13 - Prob. 26QRTCh. 13 - Refer to Figure 13.10 ( Sec. 13-4b) to answer...Ch. 13 - Prob. 28QRTCh. 13 - Prob. 29QRTCh. 13 - Prob. 30QRTCh. 13 - The Henrys law constant for nitrogen in blood...Ch. 13 - Prob. 32QRTCh. 13 - Prob. 33QRTCh. 13 - Prob. 34QRTCh. 13 - Prob. 35QRTCh. 13 - Prob. 36QRTCh. 13 - Prob. 37QRTCh. 13 - Prob. 38QRTCh. 13 - Prob. 39QRTCh. 13 - Prob. 40QRTCh. 13 - A sample of water contains 0.010 ppm lead ions,...Ch. 13 - Prob. 42QRTCh. 13 - Prob. 43QRTCh. 13 - Prob. 44QRTCh. 13 - Prob. 45QRTCh. 13 - Prob. 46QRTCh. 13 - Prob. 47QRTCh. 13 - Prob. 48QRTCh. 13 - Prob. 49QRTCh. 13 - Prob. 50QRTCh. 13 - Consider a 13.0% solution of sulfuric acid,...Ch. 13 - You want to prepare a 1.0 mol/kg solution of...Ch. 13 - Prob. 53QRTCh. 13 - Prob. 54QRTCh. 13 - Prob. 55QRTCh. 13 - A 12-oz (355-mL) Pepsi contains 38.9 mg...Ch. 13 - Prob. 57QRTCh. 13 - Prob. 58QRTCh. 13 - Prob. 59QRTCh. 13 - Prob. 60QRTCh. 13 - Prob. 61QRTCh. 13 - Prob. 62QRTCh. 13 - Prob. 63QRTCh. 13 - Prob. 64QRTCh. 13 - Prob. 65QRTCh. 13 - Prob. 66QRTCh. 13 - Calculate the boiling point and the freezing point...Ch. 13 - Prob. 68QRTCh. 13 - Prob. 69QRTCh. 13 - Prob. 70QRTCh. 13 - Prob. 71QRTCh. 13 - Prob. 72QRTCh. 13 - The freezing point of p-dichlorobenzene is 53.1 C,...Ch. 13 - Prob. 74QRTCh. 13 - Prob. 75QRTCh. 13 - A 1.00 mol/kg aqueous sulfuric acid solution,...Ch. 13 - Prob. 77QRTCh. 13 - Prob. 78QRTCh. 13 - Prob. 79QRTCh. 13 - Prob. 80QRTCh. 13 - Prob. 81QRTCh. 13 - Differentiate between the dispersed phase and the...Ch. 13 - Prob. 83QRTCh. 13 - Prob. 84QRTCh. 13 - Prob. 85QRTCh. 13 - Prob. 86QRTCh. 13 - Prob. 87QRTCh. 13 - Prob. 88QRTCh. 13 - Prob. 89QRTCh. 13 - Prob. 90QRTCh. 13 - Prob. 91QRTCh. 13 - Prob. 92QRTCh. 13 - Prob. 93QRTCh. 13 - Prob. 94QRTCh. 13 - Prob. 95QRTCh. 13 - Prob. 96QRTCh. 13 - Prob. 97QRTCh. 13 - Prob. 98QRTCh. 13 - Prob. 99QRTCh. 13 - Prob. 100QRTCh. 13 - Prob. 101QRTCh. 13 - Prob. 102QRTCh. 13 - In The Rime of the Ancient Mariner the poet Samuel...Ch. 13 - Prob. 104QRTCh. 13 - Prob. 105QRTCh. 13 - Calculate the molality of a solution made by...Ch. 13 - Prob. 107QRTCh. 13 - Prob. 108QRTCh. 13 - Prob. 109QRTCh. 13 - Prob. 110QRTCh. 13 - The organic salt [(C4H9)4N][ClO4] consists of the...Ch. 13 - A solution, prepared by dissolving 9.41 g NaHSO3...Ch. 13 - A 0.250-M sodium sulfate solution is added to a...Ch. 13 - Prob. 114QRTCh. 13 - Prob. 115QRTCh. 13 - Prob. 116QRTCh. 13 - Prob. 117QRTCh. 13 - Prob. 118QRTCh. 13 - Prob. 119QRTCh. 13 - Refer to Figure 13.10 ( Sec. 13-4b) to determine...Ch. 13 - Prob. 121QRTCh. 13 - Prob. 122QRTCh. 13 - Prob. 123QRTCh. 13 - Prob. 124QRTCh. 13 - In your own words, explain why (a) seawater has a...Ch. 13 - Prob. 126QRTCh. 13 - Prob. 127QRTCh. 13 - Prob. 128QRTCh. 13 - Prob. 129QRTCh. 13 - Prob. 130QRTCh. 13 - Prob. 131QRTCh. 13 - A 0.109 mol/kg aqueous solution of formic...Ch. 13 - Prob. 133QRTCh. 13 - Maple syrup sap is 3% sugar (sucrose) and 97%...Ch. 13 - Prob. 137QRTCh. 13 - Prob. 13.ACPCh. 13 - Prob. 13.BCPCh. 13 - Prob. 13.CCP
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