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