How much Tris do you need to weigh out to make 250 ml of the 50 mM buffer?
A. 0.95g
B. 1.51g
C. 2.42
D. 1.75g
mols = M x L - ?
Then mol = grams/molar mass
1.51g
To calculate the amount of Tris needed to make a 50 mM buffer solution, we need to know the molar mass of Tris and the volume of the buffer solution we want to prepare.
1. First, let's calculate the molar mass of Tris (Tris(hydroxymethyl)aminomethane). The molar mass of Tris is approximately 121.14 g/mol.
2. Next, we need to convert the desired concentration from millimoles per liter (mM) to moles per liter (M). Since 1 M = 1000 mM, the 50 mM buffer concentration is equivalent to 0.05 M.
3. To determine the number of moles of Tris required for the buffer solution, we multiply the volume (in liters) by the concentration (in moles per liter). In this case, the volume is 250 ml, which can be converted to liters by dividing by 1000 (1 ml = 0.001 L). Therefore, the volume is 0.25 L.
Moles of Tris = Volume (L) × Concentration (M)
= 0.25 L × 0.05 M
= 0.0125 moles
4. Finally, to convert moles to grams, we multiply the moles by the molar mass of Tris.
Mass in grams = Moles × Molar mass
= 0.0125 moles × 121.14 g/mol
≈ 1.5135 grams
Therefore, the correct answer to the question is B. 1.51g.