In order to prepare 50.0ml of 0.100M C12H22O11 (sucrose) you will add___grams of C12H22O11 to____ml of water.
Can someone explain how I would solve this? I am lost and do not understand how to do the conversions
First, the instructor that made up this problem gets a failing grade. You can not add grams sucrose to ANY amount of water. To make a molar solution you add x grams of the solute to some water, swirl until dissolved, then make to the mark on the volumetric flask. Here is what you do for 50.00 mL of a 0.1 M solution of sucrose.
How many mols of sucrose do you want? That's M x L = 0.100 M x 0.05000 L= 0.005000,
Then grams = mols x molar mass = 0.005000 x approx 342 = ? grams sucrose. You can calculate the molar mass of sucrose more accurately than that and redo the other calculations. To finish, you weigh out the sucrose, place all of it in a 50 mL volumetric flask, add some distilled or deionized water, swirl until all of the sucrose is dissolved, add distilled or deionized water to the mark on the neck of the volumetric flask, mix thoroughly, stopper, label.
The reason you don't want to add the sucrose to the flask, then add 50 mL water is because the volume will now be larger than 50 mL and the M will NOT be 0.100 M.
To solve this problem, you need to calculate the amount of C12H22O11 (sucrose) that you need to add to a certain volume of water to prepare a 0.100M solution. Here's a step-by-step guide:
Step 1: Determine the molar mass of C12H22O11.
The molar mass of C12H22O11 can be calculated by adding up the atomic masses of the constituent elements. The atomic mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol.
Molar mass of C12H22O11 = (12.01 × 12) + (1.01 × 22) + (16.00 × 11) = 342.30 g/mol
Step 2: Calculate the number of moles of C12H22O11 needed.
The molarity (M) is defined as moles of solute per liter of solution.
Molarity (M) = moles of solute / volume of solution (in liters)
0.100 M = x moles / 0.050 L (Since the volume is given as 50.0 ml, convert it to liters by dividing by 1000.)
0.100 M × 0.050 L = x moles
x = 0.0050 moles
Step 3: Convert moles to grams.
The number of moles obtained in Step 2 can be used to determine the mass of C12H22O11 needed.
Mass = number of moles × molar mass
Mass = 0.0050 moles × 342.30 g/mol = 1.71 grams
Therefore, to prepare 50.0 ml of 0.100M C12H22O11 solution, you need to add 1.71 grams of C12H22O11 to the water.
To solve this problem, you need to use the given information to find the mass of sucrose (C12H22O11) and the volume of water needed to prepare a 0.100M solution.
Step 1: Determine the molar mass of sucrose (C12H22O11).
The molar mass can be calculated by adding up the atomic masses of all the elements in a molecule. In this case, C12H22O11 contains 12 carbon (C) atoms with a molar mass of 12.01 g/mol, 22 hydrogen (H) atoms with a molar mass of 1.01 g/mol, and 11 oxygen (O) atoms with a molar mass of 16.00 g/mol.
Molar mass of C12H22O11 = (12 x 12.01) + (22 x 1.01) + (11 x 16.00) g/mol
Step 2: Calculate the moles of sucrose needed.
Given that the final volume of the solution is 50.0 mL (or 0.050 L) and the required concentration is 0.100M, we can use the formula:
moles of solute (C12H22O11) = concentration (M) x volume (L)
moles of C12H22O11 = 0.100 mol/L x 0.050 L
Step 3: Convert moles of sucrose to grams.
To convert moles to grams, you multiply the number of moles by the molar mass of the substance.
grams of C12H22O11 = moles of C12H22O11 x molar mass of C12H22O11
Step 4: Determine the volume of water needed.
The volume of water needed to prepare the solution is the final volume minus the volume of the solute added. So, the volume of water = 50.0 mL - volume of sucrose (C12H22O11) added.
Now, plug in the values into the equations and calculate the answer.