How many grams of table sugar (C12H22O11) will be needed to make 3.5 L of a 11.5 M solution?

Well, if you're looking to make a 11.5 M solution, I hope you have a sweet tooth! Now, let's do some sugar math, shall we?

First, we need to convert 3.5 L of solution to grams. The density of water is approximately 1 g/mL, which means 3.5 L of water would be 3,500 grams.

Next, we need to determine the molar mass of table sugar (C12H22O11). By adding up the atomic masses of carbon (C), hydrogen (H), and oxygen (O), we find that the molar mass of table sugar is approximately 342 g/mol.

Now, we can use the formula:

moles = concentration × volume

Since our concentration is 11.5 M and our volume is 3.5 L (or 3,500 mL), we can calculate:

moles of sugar = 11.5 M × 3.5 L = 40,250 moles

Finally, we can multiply the moles of sugar by the molar mass to find the number of grams needed:

grams of sugar = 40,250 moles × 342 g/mol ≈ 13,782,500 grams

So, you'll need approximately 13,782,500 grams of table sugar to make 3.5 L of a 11.5 M solution. That's enough sweetness to keep a clown like me going for a long time! Enjoy!

To determine the grams of table sugar (C12H22O11) needed to make the solution, you need to use the formula:

moles of solute = molarity × volume of solution

First, let's convert the volume of the solution from liters to milliliters since the molarity is given in moles per liter:

3.5 L × 1000 mL/L = 3500 mL

Next, we need to calculate the moles of table sugar:

moles of solute = molarity × volume of solution
moles of solute = 11.5 M × 3500 mL / 1000 mL/L
moles of solute = 40.25 mol

Now we need to determine the molar mass of table sugar (C12H22O11):

molar mass of C = 12.01 g/mol
molar mass of H = 1.01 g/mol
molar mass of O = 16.00 g/mol

molar mass of C12H22O11 = 12.01 g/mol × 12 + 1.01 g/mol × 22 + 16.00 g/mol × 11
molar mass of C12H22O11 = 342.34 g/mol

Lastly, we can calculate the grams of table sugar needed:

grams of table sugar = moles of solute × molar mass of C12H22O11
grams of table sugar = 40.25 mol × 342.34 g/mol
grams of table sugar = 13,785.34 g

Therefore, you would need approximately 13,785.34 grams of table sugar to make 3.5 L of a 11.5 M solution.

To find out how many grams of table sugar (C12H22O11) will be needed to make 3.5 L of an 11.5 M solution, we need to use some calculations.

1. First, let's find the number of moles of table sugar required:
The molarity (M) of a solution is defined as moles of solute per liter of solution. Therefore, Molarity (M) = moles of solute / volume of solution in liters.

Rearranging the formula, moles of solute = Molarity (M) × volume of solution in liters.

In this case, the volume is given as 3.5 L, and the molarity is 11.5 M.

Moles of table sugar = 11.5 M × 3.5 L

2. Next, we'll use the molar mass of table sugar to convert moles into grams. The molar mass of table sugar (C12H22O11) can be calculated by summing up the atomic masses of its constituent elements from the periodic table.

Carbon (C) atomic mass = 12.01 g/mol
Hydrogen (H) atomic mass = 1.01 g/mol
Oxygen (O) atomic mass = 16.00 g/mol

Molar mass of table sugar = (12 × 12.01) + (22 × 1.01) + (11 × 16.00) g/mol

3. Finally, using the molar mass, we can calculate the grams of table sugar required by multiplying the moles of table sugar by the molar mass.

Grams of table sugar = Moles of table sugar × Molar mass of table sugar

Now, let's plug in the numbers and calculate:

Moles of table sugar = 11.5 M × 3.5 L

Molar mass of table sugar = (12 × 12.01) + (22 × 1.01) + (11 × 16.00) g/mol

Grams of table sugar = Moles of table sugar × Molar mass of table sugar

By using these calculations, we can determine the exact amount of grams of table sugar needed to make 3.5 L of an 11.5 M solution.

You want how many mols? That's M x L = mols.

Then mols = grams/molar mass. YOu know mols and molar mass, solve for grams.