Can someone help with this question?
Here are the formula weights for three compounds often used in molecular biology: NaCl, 58.44; Tris base, 121.1; EDTA, 372.2 a) For each of these three compounds, how much solute is required to make 1L of a 1M solution?
b) For each of these three compounds, how much solute is required to make 0.5M of a 1L solution?
1M=molesolute/volume
moles solute=1*1=1=1 formula weigh
b. 1/2 formula weight.
To calculate the amount of solute required to make a 1M solution of each compound, we can use the formula:
Amount of solute (in grams) = Concentration (in moles/liter) x Formula weight (in grams/mole) x Volume of solution (in liters)
In this case, we want to make a 1M solution with a volume of 1 liter. Let's calculate the amount of solute required for each compound:
a) NaCl:
Concentration = 1M
Volume of solution = 1 liter
Formula weight of NaCl = 58.44 g/mol
Using the formula:
Amount of solute (in grams) = 1M x 58.44 g/mol x 1 liter = 58.44 grams of NaCl
Therefore, you would need 58.44 grams of NaCl to make a 1M solution in 1 liter.
b) Tris base:
Concentration = 1M
Volume of solution = 1 liter
Formula weight of Tris base = 121.1 g/mol
Using the formula:
Amount of solute (in grams) = 1M x 121.1 g/mol x 1 liter = 121.1 grams of Tris base
Therefore, you would need 121.1 grams of Tris base to make a 1M solution in 1 liter.
c) EDTA:
Concentration = 1M
Volume of solution = 1 liter
Formula weight of EDTA = 372.2 g/mol
Using the formula:
Amount of solute (in grams) = 1M x 372.2 g/mol x 1 liter = 372.2 grams of EDTA
Therefore, you would need 372.2 grams of EDTA to make a 1M solution in 1 liter.