How many grams of water are required to dissolve 27.8g of NH4(NO3)(MM=80.05 g/mol)to make a 0.500M solution?
I don't know. The definition of molarity = M is #moles/L of SOLUTION and not grams of water.
This is a verbatim question on my final review for 2nd quarter College General Chemistry and it didn't make any sense to me so that is why I decided to post and ask for assistance. You not knowing either doesn't make me feel any more confident in my ability to pass this class :(
To calculate the number of grams of water required to dissolve 27.8g of NH4(NO3) and make a 0.500M solution, we need to consider the concept of molarity and the molecular weight of NH4(NO3).
First, let's calculate the number of moles of NH4(NO3) using its given mass:
Number of moles = Mass / Molecular weight
Number of moles = 27.8g / 80.05 g/mol
Next, we need to calculate the volume of the solution required to get a 0.500M concentration using the equation for molarity:
Molarity = Number of moles / Volume (in liters)
Since we want to know the volume in liters, we can rearrange this equation:
Volume (in liters) = Number of moles / Molarity
Now that we have the volume required, we can calculate the mass of water. The density of water is approximately 1g/mL at room temperature, which is equivalent to 1g/cm³ or 1000g/L. Therefore, in 1 liter of water, there are 1000 grams.
Mass of water = Volume (in liters) x Density of water
Mass of water = Volume (in liters) x 1000g/L
Now let's put all the calculations together:
1. Calculate the number of moles of NH4(NO3):
Number of moles = 27.8g / 80.05 g/mol
2. Calculate the volume of the solution required:
Volume (in liters) = Number of moles / Molarity
3. Calculate the mass of water:
Mass of water = Volume (in liters) x 1000g/L
Finally, perform these calculations using the given values to find the mass of water required to dissolve 27.8g of NH4(NO3) and make a 0.500M solution.