Question

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

Explanation:

Step1: Calculate moles of KNO₃

The molar mass of KNO₃ ($K = 39.1\ g/mol$, $N=14.0\ g/mol$, $O = 16.0\ g/mol$) is $M = 39.1+14.0 + 3\times16.0=101.1\ g/mol$. The number of moles of KNO₃, $n_{KNO_3}=\frac{m}{M}=\frac{5.00\ g}{101.1\ g/mol}= 0.0495\ mol$.

Step2: Calculate molarity (a)

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 $M=\frac{0.0495\ mol}{0.1600\ L}=0.309\ M$.

Step3: Calculate % mass (b)

The total mass of the solution $m_{total}=5.00\ g + 150.0\ g=155.0\ g$. The $\%$ mass of KNO₃ is $\frac{m_{KNO_3}}{m_{total}}\times100=\frac{5.00\ g}{155.0\ g}\times100 = 3.23\%$.

Step4: Calculate molality (c)

Molality $m=\frac{n_{solute}}{m_{solvent\ in\ kg}}$. The mass of ethanol (solvent) $m_{solvent}=150.0\ g = 0.1500\ kg$. So $m=\frac{0.0495\ mol}{0.1500\ kg}=0.330\ m$.

Step5: Calculate moles of ethanol

The molar mass of ethanol ($C_2H_5OH$: $C = 12.0\ g/mol$, $H=1.0\ g/mol$, $O = 16.0\ g/mol$) is $M = 2\times12.0+6\times1.0 + 16.0=46.0\ g/mol$. The number of moles of ethanol $n_{ethanol}=\frac{150.0\ g}{46.0\ g/mol}=3.261\ mol$.

Step6: Calculate mole - fraction (d)

The mole - fraction of KNO₃, $X_{KNO_3}=\frac{n_{KNO_3}}{n_{KNO_3}+n_{ethanol}}=\frac{0.0495\ mol}{0.0495\ mol + 3.261\ mol}=0.0148$.

Step7: Calculate ppm (e)

$ppm=\frac{m_{solute}}{m_{total}}\times10^6$. Since $m_{total}=155.0\ g$, $ppm=\frac{5.00\ g}{155.0\ g}\times10^6 = 32258\ ppm$.

Answer:

a. $0.309\ M$
b. $3.23\%$
c. $0.330\ m$
d. $0.0148$
e. $32258\ ppm$