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

Explanation:

5a. Molarity of the solution

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 $M_{KNO_3}=39.1 + 14.01+3\times16.00=101.11\ g/mol$. The number of moles of KNO₃, $n_{KNO_3}=\frac{m_{KNO_3}}{M_{KNO_3}}=\frac{5.00\ g}{101.11\ g/mol}=0.0495\ mol$.

Step2: Calculate molarity

Molarity $M=\frac{n}{V}$, where $n$ is the number of moles of solute and $V$ is the volume of solution in liters. $V = 160.0\ mL=0.1600\ L$. $M=\frac{0.0495\ mol}{0.1600\ L}=0.309\ M$.

5b. % mass of KNO₃ in the solution

Step1: Calculate total mass of 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 % mass

$% \text{ mass}=\frac{m_{KNO_3}}{m_{total}}\times100%=\frac{5.00\ g}{155.0\ g}\times100% = 3.23%$.

5c. Molality of the solution

Step1: Recall molality formula

Molality $m=\frac{n_{solute}}{m_{solvent}(kg)}$. We know $n_{KNO_3}=0.0495\ mol$ and $m_{ethanol}=150.0\ g = 0.1500\ kg$. $m=\frac{0.0495\ mol}{0.1500\ kg}=0.330\ m$.

5d. Mole - fraction of KNO₃ in the solution

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 $M_{ethanol}=2\times12.01 + 6\times1.01+16.00 = 46.07\ g/mol$. The number of moles of ethanol $n_{ethanol}=\frac{m_{ethanol}}{M_{ethanol}}=\frac{150.0\ g}{46.07\ g/mol}=3.256\ mol$.

Step2: Calculate mole - fraction

The mole - fraction of KNO₃, $x_{KNO_3}=\frac{n_{KNO_3}}{n_{KNO_3}+n_{ethanol}}=\frac{0.0495\ mol}{0.0495\ mol + 3.256\ mol}=0.0148$.

5e. Parts per million (ppm) of KNO₃ in the solution

Step1: Recall ppm formula

$ppm=\frac{m_{solute}}{m_{total}}\times10^6$. We know $m_{KNO_3}=5.00\ g$ and $m_{total}=155.0\ g$. $ppm=\frac{5.00\ g}{155.0\ g}\times10^6 = 32258\ ppm$.

6a. Molar mass of the metal hydroxide

Step1: Write the balanced acid - base reaction

Let the metal hydroxide be $M(OH)_2$. The reaction with HCl is $M(OH)2+2HCl = MCl_2 + 2H_2O$. The number of moles of HCl, $n{HCl}=M\times V=2.50\ mol/L\times0.0569\ L = 0.14225\ mol$. From the stoichiometry of the reaction, the number of moles of $M(OH)2$, $n{M(OH)2}=\frac{1}{2}n{HCl}=\frac{1}{2}\times0.14225\ mol = 0.071125\ mol$. The molar mass of $M(OH)2$, $M{M(OH)2}=\frac{m{M(OH)2}}{n{M(OH)_2}}=\frac{8.65\ g}{0.071125\ mol}=121.6\ g/mol$.

6b. Identity of the metal cation

Step1: Calculate molar mass of the metal

The molar mass of $(OH)2$ is $2\times(16.00 + 1.01)=34.02\ g/mol$. The molar mass of the metal $M = M{M(OH)_2}-34.02\ g/mol=121.6\ g/mol - 34.02\ g/mol = 87.6\ g/mol$. The metal is $Sr^{2 + }$ since the molar mass of Sr is approximately $87.62\ g/mol$.

7. Concentration of $Mn^{2+}$ in the sample

Step1: Balance the redox reaction in basic medium

The unbalanced reaction is $Mn^{2+}(aq)+MnO_4^{-}(aq)\to MnO_2(s)$. In basic medium: $3Mn^{2+}(aq)+2MnO_4^{-}(aq)+2OH^{-}(aq)\to5MnO_2(s)+H_2O(l)$.

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

$n_{MnO_4^{-}}=M\times V=0.04162\ mol/L\times0.03721\ L = 0.001549\ mol$.

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

From the balanced reaction, the mole - ratio of $Mn^{2+}$ to $MnO_4^{-}$ is $3:2$. So $n_{Mn^{2+}}=\frac{3}{2}n_{MnO_4^{-}}=\frac{3}{2}\times0.001549\ mol=0.002324\ mol$.

Step4: Calculate concentration of $Mn^{2+}$

The volume of the $Mn^{2+}$ sample is $V_{Mn^{2+}}=0.0250\ L$. The concentration of $Mn^{2+}$, $M_{Mn^{2+}}=\frac{n_{Mn^{2+}}}{V_{Mn^{2+}}}=\frac{0.002324\ mol}{0.0250\ L}=0.09296\ M$.

Answer:

5a. $0.309\ M$ 5b. $3.23%$ 5c. $0.330\ m$ 5d. $0.0148$ 5e. $32258\ ppm$ 6a. $121.6\ g/mol$ 6b. $Sr^{2+}$ 7. $0.09296\ M$