uP tLSS UN MONDAY/TUESDAY, Jan30/Jan31 SHOW ALL WORK AND USE CORRECT SCIENTIFIC
ID: 575314 • Letter: U
Question
uP tLSS UN MONDAY/TUESDAY, Jan30/Jan31 SHOW ALL WORK AND USE CORRECT SCIENTIFIC NOTATION AND SIGNIFICANT FIGURES Solution problems: Calculate the amount of stock reagentfs) and water required to make the following solutions (2 points each): 1. 600mL of 45mM Tris-HCI from 1.5M stock: 1.5L of 55mM NaCl (from 1.0M stock), 25mM EDTA (from 0.5M stock) (hint: both the NaCI and the EDTA need to be put in the same bottle/beaker with the appropriate amount of water) 3. glycerol (from pure liquid stock): 4. Calculate the molarity of a solution prepared by dissolving 10.0 g of solid NaOH in enough water to make 250 mL of solution.Explanation / Answer
1) 600 mL of 45 mM Tris-HCl from 1.5 M stock solution,
Here, we can use the relation,
M1 V1=M2 V2
where, M1= Concentration of Tris required = 45 mM
V1 = volume of tris required =600 mL
M2 = concentration of stock solution =1.5 M =1500 mM
V2= Required volume of stock solution (unknown)
Therefore,
V2= (M1 V1) /M2
= (600 x 45) /1500 = 18 mL
18 mL of stock should be dissolved in 600 mL
2)
a) 1.5 L of 55 mM NaCl from 1 M stock
M1 V1=M2 V2
where, M1= Concentration of NaCl required = 55 mM
V1 = volume of NaCl required =1.5 L= 1500 mL
M2 = concentration of stock solution =1M =1000 mM
V2= Required volume of stock solution(unknown)
Therefore,
V2= (M1 V1) /M2
=(1500 x 55) /1000 = 82.5 mL
82.5 mL of stock should be dissolved in 1.5 L
b) 25 mM EDTA from 0.5M stock
Here, since both NaCl and EDTA required to be put in to same bottle with appropriate amount of water, we can assume that the volume required is 1.5 L
M1 V1 = M2 V2
where, M1= Concentration of EDTA required = 25 mM
V1 = volume of EDTA required =1.5 L= 1500 mL
M2 = concentration of stock solution =0.5 M = 500 mM
V2 = Required volume of stock solution(unknown)
Therefore,
V2 = (M1V1) /M2
=(1500 x 25) /500 = 75 mL
75 mLstock should be dissolved in 1.5 L
3)
a) 650 mL of 8 mM sodium citrate
Molar mass of sodium citrate = 258.06 g/mol
Mass of sodium citrate required to make a 650 mL 8mM solution = (0.008 /1L) x 0.650 x 258.06 =
1.341912 g
1.341912 g of sodium citrate should be dissolved in 650 mL to make 8mM solution
b) 45 mM SDS from 55 mM stock
Let us make 100 mL 45 mM SDS
We have,
M1 V1 = M2 V2
where, M1= Concentration of SDS required = 45 mM
V1 = volume of SDS required = 100 mL
M2 = concentration of stock solution =55 mM
V2= Required volume of stock solution(unknown)
Therefore,
V2= (M1 V1) /M2
= (100 x 45) /55 = 81.88 mL
81.88 mL of stock should be diluted to 100 mL
c) 0.45% glycerol
For making a 100 mL 0.45% glycerol, we will have to dissolve 0.45 g of glycerol in 100 mL water.
We have, Density of glycerol = 1.26 g/cm³
Volume of glycerol required to make a 100 mL 0.45% solution = 0.45/1.26 = 0.368 mL
4) We have ,
Molarity = (mass/molar mass) x (1000/required volume)
= (10/40) x (1000/250) = 1 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.