The molarity of the stock solution is 1.25 M.
<u>Explanation:</u>
We have to find the molarity of the stock solution using the law of volumetric analysis as,
V1M1 = V2M2
V1 = 150 ml
M1 = 0.5 M
V2 = 60 ml
M2 = ?
The above equation can be rearranged to get M2 as,
M2 = 
Plugin the values as,
M2 = 
= 1.25 M
So the molarity of the stock solution is 1.25 M.
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
Salt lowers the freezing/melting point of water, so in both cases the idea is to take advantage of the lower melting point. Ice forms when the temperature of water reaches 32 degrees Fahrenheit (0 degrees Celsius).
Answer : The freezing point of the solution is, 260.503 K
Solution : Given,
Mass of methanol (solute) = 215 g
Mass of water (solvent) = 1000 g = 1 kg (1 kg = 1000 g)
Freezing depression constant = 
Formula used :
where,
= freezing point of water = 
= freezing point of solution
= freezing point constant
= mass of solute
= mass of solvent
= molar mass of solute
Now put all the given values in the above formula, we get
By rearranging the terms, we get the freezing point of solution.
Therefore, the freezing point of the solution is, 260.503 K
B) brings order to the elements. This is correct because elements are in order by highest atomic number.
