If 25 ml of bromine, a liquid element at room temperature, is added to 100 ml of water, what would be the resulting molar concentration of bromine?
Technically I don't think this problem can be solved. We don't know the final volume of the solution and we don't know if all of the Br2 will dissolve in the 100 mL water.
To calculate the molar concentration of bromine in the resulting solution when 25 ml of bromine is added to 100 ml of water, you need to know the molar mass of bromine.
First, find the molar mass of bromine. The molar mass of bromine (Br) is approximately 79.90 grams per mole (g/mol).
Next, convert the volume of bromine from milliliters (ml) to liters (L), since molar concentration is typically expressed in moles per liter (mol/L).
25 ml of bromine is equal to 0.025 L (since 1 L = 1000 ml).
Now you can calculate the number of moles of bromine by using the formula:
moles = volume (in liters) * molarity
Since the volume of bromine is 0.025 L and you don't know the molarity yet, you'll need to rearrange the formula to solve for molarity:
molarity = moles / volume (in liters)
The number of moles of bromine is determined by dividing the given mass (which we need to calculate) by the molar mass of bromine.
To find the mass of bromine, you need to know the density of bromine. Let's assume the density of bromine is 3.12 g/ml.
The mass of bromine is calculated using the formula:
mass = volume * density
mass = 25 ml * 3.12 g/ml = 78 g
Now that you have the mass of bromine, you can calculate the moles of bromine:
moles = mass / molar mass
moles = 78 g / 79.90 g/mol = 0.976 moles
Finally, substitute the calculated moles (0.976 moles) and volume (0.025 L) into the molarity formula to find the molar concentration of bromine:
molarity = moles / volume (in liters)
molarity = 0.976 moles / 0.025 L = 39.04 mol/L
Therefore, the resulting molar concentration of bromine in the solution would be approximately 39.04 mol/L.