I have worked part a but I have no clue how to do the rest will you please show
ID: 320927 • Letter: I
Question
I have worked part a but I have no clue how to do the rest will you please show how to do them 5.) Dilutions a. You find a solution of 255 mg/mL cuso4. Is this the same as a 20% (w/v)solution? If not, how would you alter the solution to arrive at the desired concentration? b. You have a solution of 2M c6H1206. Is this the same as a 10% (w/v) solution? If not, how would you alter the solution to arrive at the desired concentration? c: You have a solution of 25 mM NaN3. ls this the same as a 15 ng/mL solution? If not, how would you alter the solution to arrive at the desired concentration? If not, how d. You have a solution of 4.59tchCIN. s this the same as a 150 mM solution would you alter the solution to arrive at the desired concentration? e. You find a solution of 103 ng/uL c21H20BrN3. Is this the same as an 8o mM solution? If not, how would you alter the solution to arrive at the desired concentration?Explanation / Answer
b) Molecular weight of C6H12O6 = 180.1559 g/mol
% w/v = (mass solute in g / volume of solution in ml) x 100
10 % w/v means 10g/100 ml
Molarity is expressed as M or mol/L
We have 2M solution which means 2mol/L
2 mol/L x 180.1559g/mol x 1L/103 ml = 0.36 g/ml
To express in %w/v, multiply numerator and denominator by 100
36.03 g/ 100ml = 36.03% w/v
So 2M is not equal to 10% w/v
To get the desired concentration,
v1 x c1 = v2 x c2
v1 x 10 = 100 x 36.03
v1 = 360.3 ml
Add 360.3 ml of solvent to 2M solution to get 10% w/v
c) Molecular weight of NaN3 = 65.0099 g/mol
25 x 10-3 mol/L x 65.0099g/mol x L/103 ml = 1.625 x 10-3 g/ml = 1.625 mg/ml
So 25mM is not equal to 15ng/ml
To get the desired concentration
C1 x v1 = c2 x v2
15 x 10-9 g x v1 = 1.625 x 10-3 g x ml
v1 = 108.33 x 103 ml = 108.33 liter
Add 108.33 liter solvent to the 25mM solution to get the desired concentration of 15ng/ml