1) Concentrated acetic acid has a molarity of 17.4. Describe the preparation of

ID: 835670 • Letter: 1

Question

1) Concentrated acetic acid has a molarity of 17.4. Describe the preparation of 400 mL of 0.35 M acetic acid solution.

2) How much 0.400 M acetic acid can be made from 5.00 mL of 20.0 M acetic acid solution?

3) A stock solution of 300 mM glycine was given to you. Describe the preparation of 5.0 mL solution containing 30 mM glycine.

4) how many grams of solid NaOH are required to prepare 500 mL of a 0.040 M solution? What is the concentration of this solution (express as % w/v)?

5) how many millilitres of 9 M NaCl are required to prepare 3.0 L of 0.003 M NaCl? express the final NaCl concentration in therms of:

(a) millimolar

(b) micromolar

Molarity of conc acetic acid= 17.4M

TO PREPARE 400ML 0F .35M

M1V1=M2V2

17.4*V1=.35*400

V1=8.04 ML

SO TO PREPARE 400ML OF .35M SOLN WE USE 8.04 ML OF ACETIC ACID OG 17.4M AND 391.95ML OF WATER

2) AMOUNT OF .4 M ACETIC ACID FORMED

M1V1=M2V2

.4* V2=5*20

AMOUNT= 250 ML CAN BE FORMED

3) AGAIN

USING ABOVE EQN

STOCK SOLN=300Mm

To prepare= 5ml OF 30Mm

300*V2=5*30

V2=.5ml

So to prepare 5ml of 30milimolar soln we take .5ml of 300milomolar soln and 4.5mi of water

4) amount of solid NaoH required=molarity*vol*molecular mass/1000l

=.04*500*40/1000 = .8 gm

5) again using the eqn used in frst soln

9*v1=3000ml*.003

V1= 1ml so 1ml 0f 9M NacL is required

Conc of NaCl in milimolar= 9*10-3

Conc of NaCl in micromolar= 9*10-6

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.