In: Chemistry
How many grams of solute are in each of the following solutions?
0.437 L of a 0.350 M K2CO3 solution =
15.5 mL of a 2.50 M AgNO3 solution =
25.2 mL of a 6.00 M H3PO4 solution =
Molarity of K2CO3 = Number of moles of K2CO3 / Volume of solution in L
0.350 = Number of moles of k2CO3 / 0.437 L
Number of moles of K2CO3 = 0.350 * 0.437 = 0.1530 mole
Number of moles of K2CO3 = mass of K2CO3 / molar mass of K2CO3
0.1530 mole = mass of K2CO3 / 138.205 g.mol^-1
Mass of K2CO3 = 0.1530 * 138.205 = 21.15 g