Question

prepare 500 ml of a reaction buffer containing .075% MgCl2, 200mM glucose, 900nM ATP, and 100mM Tris-Cl, pH 7.5

stock solutions previously prepared:

500 ml of 1.5 M glucose--- 135.12 grams of glucose needed

2 liters of 1.8 M Tirs buffer at pH7.5-------- 87.5234 grams base and 252.56 grams acid

Answer 1

Initially you have to dissolve the MgCl2 in water , add the glucose solution, add the atp (ATP is sold as a disosium salt, add the buffer and complete with tridestiled water until 500 ml.

Let's see the calculations:

For magnesiun chloride

Magnesium chloride hexahydrate Molar mass = 203.30 g/mol

mass of magnesium chloride = 500 ml x 0.075/100 = 0.375 g of MgCl2

but we need the mass of the hydrated salt = 0.375 g x 203.31/95.211 = 0.800g ( 95.211 is the molar mass of MgCl2 anhydre)

For glucose

200 mM x 0.5 L = 100 mmoles

volumen needed= 100x 10^-3 moles /1.5 mol/L = 0.0666 L = 6.66 mL

For ATP

starting from the disodium salt of ATP Molar mass=  551.14 g/mol

moles of ATP = 900 x10^-3 moles /L x 0.5 L = 1800x10^-9 moles

mass of ATP = 551.14 g/mol x 1800x10^-9 moles = 0.99 mg

For TRIS

M1V1 = M2V2 solving for V1   

V1= volume of the concentrated solution 1.8 M = M2V2/M1 = 100x10^-3Mx0.5L / 1.8M = 0.0277 L = 2.77 mL

Preparation

Add to a volumetric flask of 500 ml 0.800g of MgCl2 hexahydrated, add 6.66 mL of glucose 1.5 M solution, add 1 mg of ATP disodium salt, and add 2.77 mL of Buffer solution. Add water and check the pH, adjust with HCl, then Add tridestiled water until 500 ml and mix thoroughly.