Answer to Question #158196 in Chemistry for Jane

Question

Explain how to prepare 800 ml of 0.78 mmol/litre NAD+ (MW 655.43) solution.

Explain how to prepare 100 ml of 0.55 mmol/litre ATP (MW 555.14) solution.

 

Describe how you would prepare 400 ml of the following solution

 

Tris HCl buffer - 100 mmol/l (MW 157.6) containing

                                                                                               

1 mmol/l magesium acetate           (MW 214.45)                      

0.78 mmol/l NAD+                            (MW 655.43)                      

0.55 mmol/l ATP                              (MW 555.14)            

Answer 1

According to the equation:

M = m / (Mr × V)

where M - molarity, m - mass, Mr - molecular weight, V - volume.

From here:

m = M × Mr × V

1. m(NAD⁺) = 0.78 mmol/l × 655.43 g/mol × 800 ml = 0.78 × 10-3 mol/l × 655.43 g/mol × 0.8 l = 0.41 g

To prepare 800 ml of 0.78 mmol/ll NAD⁺ solution, it is required to take 0.41 g of NAD and adjust water up to 800 ml.

2. m(ATP) = 0.55 mmol/l × 555.14 g/mol × 100 ml = 0.55 × 10^-3 mol/l × 555.14 g/mol × 0.1 l = 0.03 g

To prepare 100 ml of 0.55 mmol/litre ATP solution, it is required to take 0.03 g of ATP and adjust water up to 100 ml.

3. m(Tris HCl) = 100 mmol/l × 157.6 g/mol × 400 ml = 6.3 g

m(Mg acetate) = 1 mmol/ × 214.45 g/mol × 400 ml = 0.086 g

m(NAD+) = 0.78 mmol/l × 655.43 g/mol × 400 ml = 0.205 g

m(ATP) = 0.55 mmol/l × 555.14 g/mo × 400 ml = 0.122 g

To prepare 400 ml of the buffer solution, it is required to take 6.3 g of Tris HCl, 0.086 g of magnesium acetate, 0.205 g of NAD, 0.122 g of ATP and adjust with water up to 400 ml.