think about the lab procedure you just read. Label each factor below V for variable or C for constant.
2 answers:
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Explanation:
As per the given data, at a higher temperature, at , the solution will occupy a larger volume than at
.
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 , density of water=0.9982071 g/ml
Therefore, will be calculated as follows.
=
=
= 1.0017961 mol/g
At , density of water = 0.9972995 g/ml
Since, =
=
Also, =
so, 1.0017961 mol/g =
= 0.9990907 mol/L
Therefore, in 500 ml, concentration of present is calculated as follows.
=
0.9990907 mol/L =
concentration = 0.49954537 mol
Hence, mass (m'') = = 50.5056 g (as molar mass of
= 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 =
where, = density of air = 0.0012 g/ml
= density of callibration weights = 8.0g/ml
d = density of weighed object
Hence, the true mass will be calculated as follows.
True mass(m) =
true mass(m) = 50.5268 g
= 50.53 g (approx)
Thus, we can conclude that 50.53 g apparent mass of needs to be measured.