How many grams of water should you add to 31.0g of sucrose,C12H22O11, to get a 0.855 m solution?
To find out how many grams of water should be added to 31.0g of sucrose (C12H22O11) to get a 0.855 m solution, we need to understand the concept of molarity (m) and how to calculate it.
Molarity (m) is defined as the number of moles of solute (in this case, sucrose) dissolved in one liter of solution. It is usually expressed in moles per liter (mol/L).
The formula to calculate molarity is:
m = (moles of solute) / (volume of solution in liters)
To find the moles of sucrose, we can use the formula:
moles = (mass of sucrose) / (molar mass of sucrose)
The molar mass of sucrose (C12H22O11) can be calculated by adding up the atomic masses of each element:
(12.01 g/mol x 12) + (1.01 g/mol x 22) + (16.00 g/mol x 11) = 342.34 g/mol
Now we can calculate the moles of sucrose:
moles = 31.0 g / 342.34 g/mol
Next, we need to calculate the volume of the solution (sucrose + water) in liters. Since we are adding water to the sucrose, the volume will not change significantly. Therefore, we can assume that the volume of the solution is approximately equal to the volume of water added.
Finally, we can rearrange the molarity formula to solve for the volume of solution:
(volume of solution) = (moles of solute) / (molarity)
In this case, we want to find the volume of water that needs to be added, so we can rearrange the formula:
(volume of water) = (moles of sucrose) / (molarity)
Substituting the values we found earlier:
(volume of water) = (31.0 g / 342.34 g/mol) / (0.855 mol/L)
Calculating this expression will give us the volume of water required in liters. To convert it to grams, we can multiply by the density of water, which is approximately 1 g/mL.
So, the final step is to convert the volume of water to grams:
(volume of water in grams) = (volume of water in liters) x (density of water in g/mL)
Now you can plug in the values and calculate the answer.