How many grams of water should you add to 33.0 of sucrose, , to get a 0.795 solution?
To determine the number of grams of water needed to create a 0.795 M solution with 33.0 grams of sucrose, you need to use the formula for calculating molarity:
Molarity (M) = moles of solute / volume of solution in liters
First, convert the grams of sucrose to moles. The molar mass of sucrose is 342.3 g/mol:
Moles of sucrose = mass of sucrose / molar mass of sucrose
= 33.0 g / 342.3 g/mol
Next, calculate the volume of the solution in liters. The molarity equation can be rearranged as:
Volume of solution (L) = moles of solute / molarity
Since we want a 0.795 M solution, plug in the known values:
Volume of solution (L) = moles of sucrose / 0.795 M
Finally, subtract the volume of the sucrose solution from the total volume to get the volume of water needed. Assume the total volume is 1 liter:
Volume of water (L) = 1 L - volume of sucrose solution
To find the mass of water, multiply the volume of water in liters by the density of water (which is about 1 g/mL or 1 g/cm³).
Mass of water (g) = Volume of water (L) * density of water
Since the density of water is 1 g/mL or 1 g/cm³, the mass of water will be the same as the volume of water in grams.
So, to summarize:
1. Convert the grams of sucrose to moles.
2. Calculate the volume of the solution in liters using the molarity equation.
3. Subtract the volume of the sucrose solution from the total volume to get the volume of water.
4. Multiply the volume of water in liters by the density of water to get the mass of water.
Implementing these steps will give you the grams of water needed to create the desired solution.