How many grams of C6H12O6 must be added to 0.750 kg of water to make a 1.25 m solution?
m = mols/kg solvent
You have m and kg solvent; solve for mols.
Then mols = grams/molar mass. You have mols and molar mass; solve for gramws.
To find out how many grams of C6H12O6 must be added to 0.750 kg of water to make a 1.25 m solution, we need to understand a few concepts.
First, let's clarify what a 1.25 m solution means. In chemistry, m stands for molarity, which is defined as the number of moles of solute per liter of solution. So, a 1.25 m solution means there are 1.25 moles of solute in every liter of solution.
To find the moles of solute, we can use the formula:
moles = mass / molar mass
In this case, the solute is C6H12O6, which is glucose. The molar mass of glucose is 180.16 g/mol (you can find this value using the periodic table).
Now, let's calculate the moles of solute required:
moles = mass / molar mass
moles = 0.750 kg / (180.16 g/mol)
To make calculations easier, we need to convert 0.750 kg to grams:
0.750 kg = 750 g
Now we can plug in the values:
moles = 750 g / 180.16 g/mol
Calculating this expression, we find:
moles ≈ 4.16 mol
So, to make a 1.25 m solution, we need approximately 4.16 moles of C6H12O6. However, we need to convert this value in moles to grams.
To convert moles to grams, we use the equation:
grams = moles × molar mass
Plugging in the values:
grams = 4.16 mol × 180.16 g/mol
Calculating this expression, we find:
grams ≈ 750 g
Therefore, you would need approximately 750 grams of C6H12O6 to make a 1.25 m solution with 0.750 kg of water.