How do I prepare 1 L of 2 mol/L phenylmagnesium bromide?
Here are the theoretical calculations for 1 L of 2 mol/L PhMgBr.
PhBr + Mg → PhMgBr
One or two I₂ crystals
Sufficient dry THF to yield 1 L
You frequently use a large excess of Mg, up to two or three times the theoretical amount. Note: The I₂ acts as a catalyst for the reaction.
Because that is all you provided for the molarity of the solution, the calculated amounts can only have one significant figure. You will need to recalculate if you require more accuracy.
By signing up, you agree to our Terms of Service and Privacy Policy
To prepare 1 L of a 2 mol/L solution of phenylmagnesium bromide, you need to dissolve the appropriate amount of phenylmagnesium bromide (C6H5MgBr) in a solvent, typically ether or THF (tetrahydrofuran), to achieve the desired concentration.
First, calculate the molar mass of phenylmagnesium bromide:
- Magnesium (Mg): 24.305 g/mol
- Bromine (Br): 79.904 g/mol
- Phenyl group (C6H5): 6 carbon atoms (12.011 g/mol each) + 5 hydrogen atoms (1.008 g/mol each) = 78.114 g/mol
Molar mass of phenylmagnesium bromide = Mg + Br + Phenyl group = 24.305 g/mol + 79.904 g/mol + 78.114 g/mol = 182.323 g/mol
Now, to prepare a 2 mol/L solution of phenylmagnesium bromide:
-
Calculate the required mass of phenylmagnesium bromide using the formula: Mass (in grams) = molarity × volume (in liters) × molar mass = 2 mol/L × 1 L × 182.323 g/mol = 364.646 g
-
Weigh out 364.646 g of phenylmagnesium bromide.
-
Dissolve the weighed phenylmagnesium bromide in enough solvent (ether or THF) to make 1 L of solution.
-
Mix the solution thoroughly until the phenylmagnesium bromide is completely dissolved.
-
Once dissolved, the solution is now ready as a 2 mol/L phenylmagnesium bromide solution.
Ensure to follow appropriate safety procedures and protocols when handling chemicals, as phenylmagnesium bromide can react violently with water and air. Additionally, perform the preparation in a well-ventilated area and wear appropriate personal protective equipment.
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- What is the mass of #9.06xx10^23# #"sodium atoms"#?
- How many grams of solute are present in 935 mL of 0.720 M #KBr#?
- Consider the reaction: #K_2S_((aq))+Co(NO_3)_(2(aq)) -> 2KNO_(3(aq))+CoS_((s)) darr#. What volume of 0.220M #K_2S# solution is required to completely react with 160mL of 0.145M #Co(NO_3)_2#?
- If 25.0 g of #K_2CO_3#, potassium carbonate, are dissolved in #450 cm^3# of solution, what is the molarity?
- What is the concentration of a sulphuric acid solution if 25.0 mL of the 36.0% w/v acid is taken by pipette and added to 500. mL of water?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7