a)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
a)
Answer to Problem 30E
0.0200 mole of Sodium phosphate in 10.0 mL of solution.
Explanation of Solution
Record the given data
Moles of Sodium phosphate=
Volume of the solution=
Calculation for the concentration of separate ions is as follows,
The balanced equation for dissolving ions,
The molarity of ions can be calculated by the formula,
Therefore, by substituting the given info in the formula, the concentrations of separate ions can be given as,
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
b)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
b)
Answer to Problem 30E
0.300 mole of Barium nitrate in 600 mL of solution.
Explanation of Solution
Record the given data
Moles of Barium nitrate=
Volume of the solution=
Calculation for the concentration of separate ions is as follows,
The balanced equation for dissolving ions
The molarity of ions can be calculated by the formula,
Therefore, by substituting the given info in the formula, molarity can be calculated by,
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
c)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
c)
Answer to Problem 30E
1.00 g of Potassium chloride in 0.500 L of solution.
Explanation of Solution
Calculation
Record the given info
Mole of Potassium chloride=1.00 gram
Volume of solution= 0.500 L
for the concentration of individual ions is as follows,
The balanced equation can be given as,
In order to calculate molarity, grams are converted into moles by using the molar mass.
The molar mass can be calculated by sum of mass of individual elements.
Molar mass of Potassium chloride can be given as (1×40) + (1×35.5) =74.55 g/mol
Amount of Potassium chloride=
=
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
d)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
d)
Answer to Problem 30E
132 g of Ammonium sulphate in 1.50 L of solution.
Explanation of Solution
Record the given info
Mole of Ammonium sulphate=132 gram
Volume of solution=1.50L
Calculation for the concentration of individual ions is as follows,
The balanced equation for dissolving ions,
In order to calculate molarity, grams are converted into moles by using the molar mass.
The molar mass can be calculated by sum of mass of individual elements.
Molar mass of Ammonium sulphate is 132 g/mol
Therefore, from the amount of Ammonium sulphate, the molarity can be calculated as,
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
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Chapter 4 Solutions
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