What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent
1 answer:
The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
<u>Explanation:</u>
Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.
As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.
Hence,
Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
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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.
Answer: The approximate molecular mass of the polypeptide is 856 g/mol
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
Or,
where,
= osmotic pressure of the solution = 4.19 torr
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (polypeptide) = 0.327 g
Volume of solution = 1.70 L
R = Gas constant =
T = temperature of the solution =
Putting values in above equation, we get:
Hence, the molar mass of the polypeptide is 856 g/mol