A aqueous solution of acetic acid with a density of 1.25 g/ml is 30.0% by mass HC2h3O2 (MM=60.0) Compute the molarity of this solution.
I figured out that if you divide 30.0 by the molar mass 60.0g which gives you .5 mol. I cannot seem to figure out what i change 1.25 g/ml into for the molarity equation.
Take a liter of solution. How much does it weigh?
1000 mL x `1.25 g/mL = 1250 grams.
How much of that is HC2H3O2? It is 1250 x 0.30 = ??
How many moles in that?
??/molar mass HC2H3O2.
M = mols/L. You have the moles and you have it in one liter. Voila!
Calculate the molarity of a solution that contains 64.0 g Al2(SO4)3in 340.5 mL water.
To find the molarity of the solution, you first need to convert the given density into molarity units (moles per liter).
The equation to convert density into molarity is:
Molarity (M) = (Mass of solute / Molar mass of solute) / Volume of solution in liters
In this case, the molar mass of acetic acid (HC2H3O2) is 60.0 g/mol, and the density of the solution is 1.25 g/ml.
To convert the density of 1.25 g/ml into g/L, you multiply by the conversion factor 1000 ml/L:
1.25 g/ml * 1000 ml/L = 1250 g/L
Now, using the equation above, you can calculate the molarity of the acetic acid solution:
Molarity (M) = (0.30 * 1250 g/L) / 60.0 g/mol
Molarity = 6.25 mol/L
Therefore, the molarity of the solution is 6.25 mol/L.
To determine the molarity of the solution, you need to convert the given density (1.25 g/mL) into the appropriate units.
The molarity (M) is defined as moles of solute per liter of solution. Therefore, you need to convert the density into grams of acetic acid per liter of solution.
Since the density is given in grams per milliliter, you can convert it into grams per liter by multiplying by 1000 (since there are 1000 mL in one liter):
1.25 g/mL * 1000 mL/L = 1250 g/L
Now, you have the mass of acetic acid (in grams) per liter of solution.
Next, you need to convert the mass percent of acetic acid (30.0% by mass) into moles of acetic acid.
Given that the molar mass of acetic acid (HC2H3O2) is 60.0 g/mol, you correctly calculated that 30.0% by mass corresponds to 0.5 moles of acetic acid.
Finally, you can use the equation for molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
Plugging in the values:
Molarity = 0.5 mol / 1 L = 0.5 M
Therefore, the molarity of the solution is 0.5 M.