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In: Chemistry

#6... 4. How would you make 500 ml of 5 M NaCl? 5. How would you...

#6...

4. How would you make 500 ml of 5 M NaCl?

5. How would you make 1 liter of 2 M sodium citrate?

6. How would you make 1 liter of stock 20X SSC (standard sodium citrate) buffer that is 3 M NaCl, 0.3 M sodium citrate, using the solutions you made in Q4 and Q5 above?

Expert Solution

4.How would you make 500 ml of 5 M NaCl?

Answer: Molarity = (weight of Solute/Mol.weight) (1/volume in L)

5M = (weight of NaCl/ 58.44 g/mol )(1/0.5L)

Weight of NaCL required = 5 M x 0.5 L x58.44 = 146.1 g

Therefore, we need to weigh 146.1 g of NaCl and dissolve into 500 mL solution to make our required solution(500mL of 5M NaCl).

5. How would you make 1 liter of 2 M sodium citrate?

Answer : Molarity = (weight of Solute/Mol.weight) (1/volume in L)

2 M = (weight of Sodium Citrate)/ 258

Weight of Sodium Citrate= 516 g

Therefore, we need to weigh 516 g of sodium citrate and dissolve it into 1 L water. This will give us the 1L of 2M sodium citrate solution.

6. How would you make 1 liter of stock 20X SSC (standard sodium citrate) buffer that is 3 M NaCl, 0.3 M sodium citrate, using the solutions you made in Q4 and Q5 above?

20X SSC solution contains 3M NaCl and 0.3 M sodium citrate.

We need to find out how much volume of 2M sodium citrate and 5M of NaCl is required.

Let us find out for NaCl:

Initial volume V1 =?

Initial concentration M1 = 5M

Final Volume V2 = 1L

Final Concentration M2 = 3M

M1V1 = M2V2

V1= M2V2/M1 = 3M x 1L /5M = 0.6 L

For sodium citrate:

Initial volume V1 =?

Initial concentration M1 = 2M

Final Volume V2 = 1L

Final Concentration M2 = 0.3M

M1V1 = M2V2

V1= M2V2/M1 = 0.3M x 1L /2M = 0.15 L

Therefore, we need to mix 600mL of 5M NaCl and 150 mL of sodium citrate solution and make up the final volume to 1L(i.e, 250 mL of water).

queen_honey_blossom answered 2 years ago

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