Given 50g and 27ml find density
1 answer:
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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.
Alpha is the -OH group at the anomeric position is down. beta is up
a. alpha
b. beta
c. beta
d. alpha
to draw the other anomer, just flip the OH at the anomeric position
a. alpha
b. beta
c. beta
d. alpha
to draw the other anomer, just flip the OH at the anomeric position