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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?

1 Answer

5 votes

Step-by-step 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,
(concentration)/(density) will be calculated as follows.

=
(C_(1))/(d_(1))

=
(1.000 mol/L)/(0.9982071 g/ml)

= 1.0017961 mol/g

At
24^(o)C, density of water = 0.9972995 g/ml

Since,
(concentration)/(density) =
(C_(2))/(d_(2))

=
(C_(2))/(0.9972995)

Also,
(C_(1))/(d_(1)) =
(C_(2))/(d_(2))

so, 1.0017961 mol/g =
(C_(2))/(0.9972995)


C_(2) = 1.0017961 * 0.9972995

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of
KNO_(3) present is calculated as follows.


C_(2) =
(concentration)/(volume)

0.9990907 mol/L =
(concentration)/(0.5 L)

concentration = 0.49954537 mol

Hence, mass (m'') =
0.49954537 mol * 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''' * ((1 - (d_(air))/(d_(weights))))/((1 - (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 * ((1 - (0.0012)/(8.0)))/((1 - ((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.

User Mirian
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