5. a solution is prepared by dissolving 5.00 g of potassium nitrate (kno₃) in 150.0 g of ethanol (c₂h₅oh). the total volume of the solution is 160.0 ml. calculate the following:

a. molarity of the solution

b. % mass of kno₃ in the solution

c. molality of the solution

d. mole fraction of kno₃ in the solution

e. parts per million (ppm) of kno₃ in the solution

6. an 8.65 g sample of an unknown group 2a metal hydroxide is dissolved in 85.0 ml of water. an acid - base indicator is added and the resulting solution is titrated with 2.50 m hcl(aq) solution. the indicator change color signaling that the equivalence point has been reached after 56.9 ml of the hydrochloric acid solution has been added.

a. what is the molar mass of the metal hydroxide?

b. what is the identity of the metal cation: ca²⁺, sr²⁺, ba²⁺?

7. the concentration of mn²⁺(aq) can be determined by titration with mno₄⁻(aq) in basic solution. a 25.00 ml sample of mn²⁺(aq) requires 37.21 ml of 0.04162 m kmno₄(aq) for its titration. what is the concentration of mn²⁺ in the sample? (hint: balance the redox reaction below first in basic medium)
mn²⁺(aq) + mno₄⁻(aq) → mno₂(s) (not balanced)

Question

5. a solution is prepared by dissolving 5.00 g of potassium nitrate (kno₃) in 150.0 g of ethanol (c₂h₅oh). the total volume of the solution is 160.0 ml. calculate the following:

a. molarity of the solution

b. % mass of kno₃ in the solution

c. molality of the solution

d. mole fraction of kno₃ in the solution

e. parts per million (ppm) of kno₃ in the solution

6. an 8.65 g sample of an unknown group 2a metal hydroxide is dissolved in 85.0 ml of water. an acid - base indicator is added and the resulting solution is titrated with 2.50 m hcl(aq) solution. the indicator change color signaling that the equivalence point has been reached after 56.9 ml of the hydrochloric acid solution has been added.

a. what is the molar mass of the metal hydroxide?

b. what is the identity of the metal cation: ca²⁺, sr²⁺, ba²⁺?

7. the concentration of mn²⁺(aq) can be determined by titration with mno₄⁻(aq) in basic solution. a 25.00 ml sample of mn²⁺(aq) requires 37.21 ml of 0.04162 m kmno₄(aq) for its titration. what is the concentration of mn²⁺ in the sample? (hint: balance the redox reaction below first in basic medium)
mn²⁺(aq) + mno₄⁻(aq) → mno₂(s) (not balanced)

Accepted Answer

### 5a. Molarity of the solution
# Explanation:
## Step1: Calculate moles of KNO₃
The molar mass of KNO₃ ($K = 39.1\ g/mol$, $N=14.01\ g/mol$, $O = 16.00\ g/mol$) is $M = 39.1+14.01 + 3	imes16.00=101.11\ g/mol$. The number of moles of KNO₃, $n_{KNO_3}=\frac{m}{M}=\frac{5.00\ g}{101.11\ g/mol}=0.04945\ mol$.
## Step2: Calculate molarity
Molarity $M=\frac{n}{V}$, where $n$ is the number of moles of solute and $V$ is the volume of the solution in liters. $V = 160.0\ mL=0.1600\ L$. So $M=\frac{0.04945\ mol}{0.1600\ L}=0.309\ M$.
# Answer:
$0.309\ M$

### 5b. % mass of KNO₃ in the solution
# Explanation:
## Step1: Calculate the total mass of the solution
The mass of KNO₃ is $m_{KNO_3}=5.00\ g$ and the mass of ethanol is $m_{ethanol}=150.0\ g$. The total mass of the solution $m_{total}=m_{KNO_3}+m_{ethanol}=5.00\ g + 150.0\ g=155.0\ g$.
## Step2: Calculate the mass - percentage
The mass - percentage of KNO₃ is $\%m=\frac{m_{KNO_3}}{m_{total}}	imes100\%=\frac{5.00\ g}{155.0\ g}	imes100\% = 3.23\%$.
# Answer:
$3.23\%$

### 5c. Molality of the solution
# Explanation:
## Step1: Recall the molality formula
Molality $m=\frac{n}{m_{solvent}}$, where $n$ is the number of moles of solute and $m_{solvent}$ is the mass of the solvent in kilograms. We know $n_{KNO_3}=0.04945\ mol$ and $m_{ethanol}=150.0\ g = 0.1500\ kg$.
## Step2: Calculate molality
$m=\frac{0.04945\ mol}{0.1500\ kg}=0.330\ m$.
# Answer:
$0.330\ m$

### 5d. Mole fraction of KNO₃ in the solution
# Explanation:
## Step1: Calculate moles of ethanol
The molar mass of ethanol $C_2H_5OH$ ($C = 12.01\ g/mol$, $H=1.01\ g/mol$, $O = 16.00\ g/mol$) is $M = 2	imes12.01+6	imes1.01 + 16.00=46.08\ g/mol$. The number of moles of ethanol $n_{ethanol}=\frac{m}{M}=\frac{150.0\ g}{46.08\ g/mol}=3.255\ mol$.
## Step2: Calculate mole - fraction
The mole - fraction of KNO₃, $x_{KNO_3}=\frac{n_{KNO_3}}{n_{KNO_3}+n_{ethanol}}=\frac{0.04945\ mol}{0.04945\ mol + 3.255\ mol}=0.0148$.
# Answer:
$0.0148$

### 5e. Parts per million (ppm) of KNO₃ in the solution
# Explanation:
## Step1: Recall the ppm formula
$ppm=\frac{m_{solute}}{m_{total}}	imes10^6$. We know $m_{KNO_3}=5.00\ g$ and $m_{total}=155.0\ g$.
## Step2: Calculate ppm
$ppm=\frac{5.00\ g}{155.0\ g}	imes10^6=32258\ ppm$.
# Answer:
$32258\ ppm$

### 6a. Molar mass of the metal hydroxide
# Explanation:
## Step1: Write the reaction equation
The metal hydroxide of group 2A has the formula $M(OH)_2$. The reaction with HCl is $M(OH)_2+2HCl = MCl_2 + 2H_2O$.
## Step2: Calculate moles of HCl
$n_{HCl}=M	imes V$, where $M = 2.50\ M$ and $V = 56.9\ mL=0.0569\ L$. So $n_{HCl}=2.50\ mol/L	imes0.0569\ L = 0.14225\ mol$.
## Step3: Calculate moles of $M(OH)_2$
From the stoichiometry of the reaction, $n_{M(OH)_2}=\frac{1}{2}n_{HCl}=\frac{1}{2}	imes0.14225\ mol = 0.071125\ mol$.
## Step4: Calculate molar mass
The molar mass $M=\frac{m}{n}$, where $m = 8.65\ g$ and $n = 0.071125\ mol$. So $M=\frac{8.65\ g}{0.071125\ mol}=121.6\ g/mol$.
# Answer:
$121.6\ g/mol$

### 6b. Identity of the metal cation
# Explanation:
## Step1: Determine the molar mass of the metal
The molar mass of $M(OH)_2$ is $M + 2	imes(16.00 + 1.01)$. We know $M_{M(OH)_2}=121.6\ g/mol$. So $M+2	imes17.01 = 121.6\ g/mol$, then $M=121.6-34.02 = 87.6\ g/mol$.
## Step2: Identify the metal
The molar mass of $Sr$ is approximately $87.62\ g/mol$. So the metal cation is $Sr^{2 + }$.
# Answer:
$Sr^{2+}$

### 7. Concentration of $Mn^{2+}$ in the sample
# Explanation:
## Step1: Balance the redox reaction in basic medium
The balanced redox reaction is $3Mn^{2+}(aq)+2MnO_4^{-}(aq)+2OH^{-}(aq)
ightarrow5MnO_2(s)+H_2O(l)$.
## Step2: Calculate moles of $MnO_4^{-}$
$n_{MnO_4^{-}}=M	imes V$, where $M = 0.04162\ M$ and $V = 37.21\ mL = 0.03721\ L$. So $n_{MnO_4^{-}}=0.04162\ mol/L	imes0.03721\ L=0.001549\ mol$.
## Step3: Calculate moles of $Mn^{2+}$
From the stoichiometry of the balanced reaction, $n_{Mn^{2+}}=\frac{3}{2}n_{MnO_4^{-}}=\frac{3}{2}	imes0.001549\ mol = 0.002324\ mol$.
## Step4: Calculate the concentration of $Mn^{2+}$
The volume of the $Mn^{2+}$ sample is $V = 25.00\ mL=0.02500\ L$. The concentration $M=\frac{n}{V}=\frac{0.002324\ mol}{0.02500\ L}=0.09296\ M$.
# Answer:
$0.09296\ M$

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QUESTION IMAGE

Question

Explanation:

5a. Molarity of the solution

Step1: Calculate moles of KNO₃

Step2: Calculate molarity

Step1: Calculate the total mass of the solution

Step2: Calculate the mass - percentage

Step1: Recall the molality formula

Step2: Calculate molality

Answer:

5b. % mass of KNO₃ in the solution