A scientist needs to check several parameters before coming
to a conclusion about the amount of water pollution. The scientists needs to
check the amount of dissolved oxygen in the water, temperature of the water,
the clarity of the water, the PH level of the water and also the amount of
bacteria present in the water. There may be other criteria’s, but the mentioned
ones are enough to gauge the amount of pollution in the water. Scientists often
takes fish and aquatic plants from the water to be tested to check the amount
of pollution indirectly affecting these species.
I literally hate chem but I think it’s ionic, I’m not not completely sure but it kinda sounds about right . Not the best anwser haha hope it kinda helps lol
Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.

Also, 
where,
is the difference in specific volumes.
Hence, 
As,
= 22.0 J/mol K
And,
...... (1)
where,
= density of water
= density of ice
M = molar mass of water =
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
=
=
Therefore, calculate the required pressure as follows.

=
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP =
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.