C).dd offspring have a nice day
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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It will become an ion and no longer be neutral
Answer:
= 25 ppm
Explanation:
- PPM also refers to parts per million, it represents a low concentration of a solution. It represents 0.001 gram or a milligram in a 1000 mL, which equivalent to 1 mg per liter.
Given; a sample size of 2000 g contained 0.050 g DDT
It means, 2000 mL sample contained 50 mg DDT
Therefore in ppm we get;
= 50 mg/ 2 L
= 25 mg/L
<u>= 25 ppm</u>
