How many grams of solute are in 0.504 L of a 0.350 M K2CO3 solution?

mols = M x L = ?

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

To determine the number of grams of solute in a solution, you need to use the equation:

grams of solute = moles of solute x molar mass of solute

First, let's calculate the moles of solute:

moles of solute = solution volume (in liters) x molarity

Given:
Solution volume = 0.504 L
Molarity = 0.350 M

Substituting the values into the equation, we get:

moles of solute = 0.504 L x 0.350 M
moles of solute = 0.1764 mol

Next, we need to determine the molar mass of K2CO3.

The molar mass of K2CO3 can be calculated by adding up the atomic masses of each element in the compound:

(2 x atomic mass of K) + atomic mass of C + (3 x atomic mass of O)

The atomic masses are:
K = 39.10 g/mol
C = 12.01 g/mol
O = 16.00 g/mol

Substituting the values into the equation, we get:

molar mass of K2CO3 = (2 x 39.10 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol)
molar mass of K2CO3 = 138.20 g/mol + 12.01 g/mol + 48.00 g/mol
molar mass of K2CO3 = 198.21 g/mol

Now, we can calculate the grams of solute using the moles of solute and the molar mass of K2CO3:

grams of solute = moles of solute x molar mass of solute
grams of solute = 0.1764 mol x 198.21 g/mol
grams of solute = 34.95 g

Therefore, there are approximately 34.95 grams of K2CO3 in 0.504 liters of a 0.350 M K2CO3 solution.