how many grams of ZnCl2 are needed to prepare 250 ml of an aqueous solution with an osmolarity equal to 0.300 osmol?
I posted this question a few minutes ago, but forgot to add the last part of the sentence. Thanks!
I think I answered it.
Osm = i*M where i is the van't Hoff factor.
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To determine the number of grams of ZnCl2 needed to prepare a solution with a specific osmolarity, we need to use the formula:
Osmolarity = (n/V) x MW
Where:
Osmolarity is the desired osmolarity (in osmol/L)
n is the number of moles of solute
V is the volume of the solution (in L)
MW is the molar mass of the solute (in g/mol)
First, we need to rearrange the formula to solve for n (moles):
n = (Osmolarity x V) / MW
Given:
Osmolarity = 0.300 osmol/L
V = 250 mL (convert to L by dividing by 1000: 250 mL / 1000 = 0.250 L)
The molar mass of ZnCl2 can be found by summing the atomic masses of zinc (Zn) and two chlorine (Cl) atoms:
Zn: atomic mass = 65.38 g/mol
Cl: atomic mass = 35.45 g/mol
Molar mass of ZnCl2 = (1 x 65.38) + (2 x 35.45) = 136.23 g/mol
Now, substitute the given values into the formula:
n = (0.300 osmol/L x 0.250 L) / 136.23 g/mol
Calculate n:
n = 0.00736 moles
Finally, to convert moles to grams of ZnCl2, we need to multiply by the molar mass:
grams = moles x MW
grams = 0.00736 moles x 136.23 g/mol
grams = 1.003 g
Therefore, approximately 1.003 grams of ZnCl2 are needed to prepare 250 mL of an aqueous solution with an osmolarity equal to 0.300 osmol.