Question

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. How many grams of solid KNO3 (density = 2.109 g/mL) should be dissolved in a volume of 500.0 mL at 24°C to give a concentration of 1.000 M at 20°C? What apparent mass of KNO; weighed in air is required?

Answer 1

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$, the solution will occupy a larger volume than at $20^{o}C$.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$, density of water=0.9982071 g/ml  

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

                 = $\frac{C_{1}}{d_{1}}$

                 = $\frac{1.000 mol/L}{0.9982071 g/ml}$

                 = 1.0017961 mol/g  

At $24^{o}C$, density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

                           = $\frac{C_{2}}{0.9972995}$

Also,             $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so,                   1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

                      $C_{2} = 1.0017961 \times 0.9972995$

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

             $C_{2}$ = $\frac{concentration}{volume}$

               0.9990907 mol/L = $\frac{concentration}{0.5 L}$  

               concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g       (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where,          $d_{air}$ = density of air = 0.0012 g/ml

                     $d_{weight}$ = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.