I need help with this question. 1).Of the 100mL of stock solution that is to be
ID: 888260 • Letter: I
Question
I need help with this question.
1).Of the 100mL of stock solution that is to be prepared for part A.1, how many milliliters (total value) will be use for preparing the standard solution IN Part A.2.
2) A 100.0-mL volume of a 0.10 M stock solution of Cu^2 is to be prepared using CuSO4*5H2O (molar mass= 249.68 g/mol)^2.
a) How many grams of CuSO4*5H2O must be measure for the preparation?
b) Describe the produce for the preparation of the solution using 0.01 M HNO3 as a diluent, starting with conc (16M) HNO3.
c)A 2.0 mL pipet transfer the Cu^2+ stock solution to a 25.0-mL volumetric flask, witch is then diluted to the mark of the volumetric flask with 0.01 M HNO3.What is the molar concentration of the diluted Cu^2+ solution?
Thank you!
Explanation / Answer
For Question 1. I need more info about the stock solution prepared so I can solve that question. Post those data along with the question on another question so I can answer ir for you.
2. a) M = mole/V; solving for mole = MxV so:
mole = 0.1 x 0.1 = 0.01 moles
m = mole x MM = 0.01 x 249.68 = 2.4968 g need to be measure for this preparation
b) Let me tell you step by step like a cooking recipe:
1- Weight 2.4968 g of CuSO4x5H2O.
2. Put it in a beaker or flask of 100 mL.
3. Add some water to the flask, little by little, and steer the mix until the solid is disolved in the water.
4. Once the solid is disolved, add water until the mark of 100 mL. This is a 0.01 M solution of CuSO4x5H2O.
5. To dilute even more the solution, use a 0.01 M HNO3 solution (volume 100 mL).
a. With a micropipet, take about 0.0625 mL of this solution (16 M) and dilute with water until 100 mL.
6. Take 2 mL from the stock Cu+2 solution with a pipet, and transfer to another flask (25 mL).
7. Add some HNO3 to the solution and complete until the 25 mL.
3) If the solution is diluted, then: 0.1 x (2/25) = 0.008 M