Answer:
28.20 mL of the stock solution.
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 12.1 M
Volume of diluted solution (V2) = 350.0 mL
Molarity of diluted solution (M2) = 0.975 M
Volume of stock solution needed (V1) =..?
The volume of stock solution needed can be obtained by using the dilution formula as shown below:
M1V1 = M2V2
12.1 x V1 = 0.975 x 350
Divide both side by 12.1
V1 = (0.975 x 350)/12.1
V1 = 28.20 mL.
Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.
Answer:
3.50 molal
Explanation:
Molality → Moles of solute / kg of solvent.
Let's convert the solvent's mass from g to kg
16.2 g . 1kg / 1000 g = 0.0162 kg
Let's determine the moles from the solute
2.61 g . 1 mol / 46 g = 0.0567 moles
Molality → 0.0567 mol / 0.0162 kg = 3.50 m
It increases as temperature rises.
