5. a solution is prepared by dissolving 5.00 g of potassium nitrate (kno₃) in 150.0 g of ethanol (c₂h₅oh)…

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:\n\na. molarity of the solution\n\nb. % mass of kno₃ in the solution\n\nc. molality of the solution\n\nd. mole fraction of kno₃ in the solution\n\ne. parts per million (ppm) of kno₃ in the solution\n\n6. 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.\n\na. what is the molar mass of the metal hydroxide?\n\nb. what is the identity of the metal cation: ca²⁺, sr²⁺, ba²⁺?\n\n7. 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)\nmn²⁺(aq) + mno₄⁻(aq) → mno₂(s) (not balanced)

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:\n\na. molarity of the solution\n\nb. % mass of kno₃ in the solution\n\nc. molality of the solution\n\nd. mole fraction of kno₃ in the solution\n\ne. parts per million (ppm) of kno₃ in the solution\n\n6. 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.\n\na. what is the molar mass of the metal hydroxide?\n\nb. what is the identity of the metal cation: ca²⁺, sr²⁺, ba²⁺?\n\n7. 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)\nmn²⁺(aq) + mno₄⁻(aq) → mno₂(s) (not balanced)

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 $M = 39.1+14.01 + 3\times16.00=101.11\ g/mol$. 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 $%m=\frac{m_{KNO_3}}{m_{total}}\times100%=\frac{5.00\ g}{155.0\ g}\times100% = 3.23%$.

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 $M = 2\times12.01+6\times1.01 + 16.00=46.08\ g/mol$. 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

$ppm=\frac{m_{solute}}{m_{total}}\times10^6$. We know $m_{KNO_3}=5.00\ g$ and $m_{total}=155.0\ g$.

Step2: Calculate ppm

$ppm=\frac{5.00\ g}{155.0\ g}\times10^6=32258\ ppm$.

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

$n_{HCl}=M\times V$, where $M = 2.50\ M$ and $V = 56.9\ mL=0.0569\ L$. So $n_{HCl}=2.50\ mol/L\times0.0569\ L = 0.14225\ mol$.

Step3: Calculate moles of $M(OH)_2$

From the stoichiometry of the reaction, $n_{M(OH)2}=\frac{1}{2}n{HCl}=\frac{1}{2}\times0.14225\ mol = 0.071125\ mol$.

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 $M + 2\times(16.00 + 1.01)$. We know $M{M(OH)_2}=121.6\ g/mol$. So $M+2\times17.01 = 121.6\ g/mol$, 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 $3Mn^{2+}(aq)+2MnO_4^{-}(aq)+2OH^{-}(aq)\rightarrow5MnO_2(s)+H_2O(l)$.

Step2: Calculate moles of $MnO_4^{-}$

$n_{MnO_4^{-}}=M\times V$, where $M = 0.04162\ M$ and $V = 37.21\ mL = 0.03721\ L$. So $n_{MnO_4^{-}}=0.04162\ mol/L\times0.03721\ L=0.001549\ mol$.

Step3: Calculate moles of $Mn^{2+}$

From the stoichiometry of the balanced reaction, $n_{Mn^{2+}}=\frac{3}{2}n_{MnO_4^{-}}=\frac{3}{2}\times0.001549\ mol = 0.002324\ mol$.

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:

$0.09296\ M$