How many grams of KBr are contained in 200 mL of a 0.310M KBr solution?
M x L x molar mass = grams.
To determine the number of grams of KBr in a given volume of solution, we need to use the concept of molarity. Molarity (M) is defined as the number of moles of solute per liter of solution.
First, convert the volume of the solution from milliliters (mL) to liters (L):
200 mL = 200/1000 L = 0.2 L
Next, use the definition of molarity to calculate the number of moles of KBr in the solution:
Molarity (M) = Moles of solute / Liters of solution
0.310 M = Moles of KBr / 0.2 L
Rearranging the equation to solve for moles of KBr:
Moles of KBr = Molarity × Liters of solution
= 0.310 M × 0.2 L
= 0.062 moles
Now, we need to convert the moles of KBr to grams using its molar mass. The molar mass of KBr is 39.0983 g/mol (for potassium) + 79.904 g/mol (for bromine).
Molar mass of KBr = 39.0983 g/mol + 79.904 g/mol = 119.0023 g/mol
To calculate the mass of KBr in grams:
Mass of KBr = Moles of KBr × Molar mass of KBr
= 0.062 moles × 119.0023 g/mol
≈ 7.378 grams
Therefore, there are approximately 7.378 grams of KBr contained in 200 mL of a 0.310 M KBr solution.