If you have 324 grams KBr and want to produce a 1.48 M solution in water how many grams of water are needed?

Hmm, is that really capital letter 'M' (Molarity) or small letter 'm' (molality)? Anyway, I'll just assume it's Molarity.

First, we get the molar mass of KBr. You'll need a periodic table to to solve for it.
Anyway, KBr has molar mass of 119 g/mol.
then we calculate for its number of moles by dividing the given mass by the molar mass:
324 g / 119 g/mol = 2.72269 mol KBr
Then we get the total volume of the solution, from the given concentration:
Molarity = moles of solute / volume of solution (in liters)
M = n/V
V = n/M
V = 2.72268 / 1.48
V = 1.8397 L
Note that this is the total volume of solution (KBr in water), but we only need the grams of water.
What we need is the density of KBr so that we can calculate the equivalent mass for this volume. Density changes with temperature (as well as the concentration given), this problem is hard to solve if there's no density given.
But, ASSUMING a density of 1 g/mL for the solution,
d = m/V
1 = m / (1.8397 * 1000)
m = 1839.7 g
Finally we subtract the given mass of KBr by the total mass of solution:
1839.7 - 324 = 1515.7 g H2O

This problem would be easier to solve if the given concentration is in molality units (mol solute / kg solvent) rather than molarity (mol solute / L solution).
hope this helps~ `u`

I don't have any idea. If you know the density of the solution we might make some headway. But grams solvent usually is reserved for m solutions. So we could work the problem for 1.48 m. For M we can't even calculate mL H2O.

To determine the grams of water needed to make a 1.48 M solution of KBr, we need to use the formula:

Molarity (M) = moles of solute / liters of solution

First, let's calculate the moles of KBr. The molar mass of KBr is 119 g/mol:

Moles of KBr = mass of KBr / molar mass of KBr

Moles of KBr = 324 g / 119 g/mol = 2.72 mol

Next, we need to calculate the liters of solution. Since the molarity is given as 1.48 M, we can rearrange the formula to solve for liters:

Liters of solution = moles of solute / molarity

Liters of solution = 2.72 moles / 1.48 M = 1.84 L

Finally, to find the grams of water needed, we use the fact that the total volume of the solution (1.84 L) is equal to the sum of the volume of the water and the volume of the solute (KBr):

Total volume = volume of water + volume of KBr

Rearranging the equation, we can solve for the volume of water:

Volume of water = Total volume - volume of KBr

Since we want the volume in grams, we can use 1 gram of water is equal to 1 milliliter.

Volume of water = (1.84 L) - (1.84 L) = 0.84 L = 840 grams of water

Therefore, you would need 840 grams of water to produce a 1.48 M solution of KBr.