how much of each of the following substances would you add to water to prepare 1.20L of a .85M solution? (sodium carbonate from the pure solid) ? I am stuck on how to work this one out, too. thanks

You need mols Na2CO3 = L x M = 1.20 x 0.085 M = ??

mols Na2CO3 = g/molar mass
easier than the last one.

To determine how much sodium carbonate you would need to prepare a 1.20L 0.85M solution, you will first need to calculate the number of moles required. Then you can use the molar mass of sodium carbonate to convert moles into grams.

Step 1: Calculate the number of moles needed:
Molarity (M) = moles/volume (L)

Given that the desired molarity is 0.85M and the desired volume is 1.20L, we can rearrange the equation to solve for moles:

moles = Molarity * volume

moles = 0.85M * 1.20L
moles = 1.02 moles

So, you would need 1.02 moles of sodium carbonate.

Step 2: Convert moles to grams:
To convert moles to grams, you need to know the molar mass of sodium carbonate.

The molar mass of sodium carbonate (Na2CO3) can be calculated by adding up the atomic masses of each element in the compound:
Na: 22.99 g/mol (sodium)
C: 12.01 g/mol (carbon)
O: 16.00 g/mol (oxygen)

molar mass of Na2CO3 = (2 * 22.99) + (12.01) + (3 * 16.00)
molar mass of Na2CO3 = 105.99 g/mol

To find the mass in grams, multiply the number of moles by the molar mass:

mass (grams) = moles * molar mass

mass (grams) = 1.02 moles * 105.99 g/mol
mass (grams) ≈ 108.09 grams

Therefore, to prepare a 1.20L 0.85M solution of sodium carbonate, you would need to add approximately 108.09 grams of sodium carbonate to the water.