How many grams of KBr are required to make 350. mL of a 0.115 mol L^-1 KBr solution? A) 0. 338 g B) 3.04 g C) 4.79 g D) 40.3 g

Question

Quizplus · Accepted Answer

To find the grams of KBr required, we need to use the formula:

grams = moles x molar mass

First, we need to calculate the number of moles of KBr in the solution:

moles = volume (in L) x molarity
moles = 0.350 L x 0.115 mol/L
moles = 0.04025 mol

Next, we need to find the molar mass of KBr:

KBr = 1 x 39.10 g/mol (one potassium atom) + 1 x 79.90 g/mol (one bromine atom)
Molar mass of KBr = 119.00 g/mol

Finally, we can use the formula to find the grams of KBr required:

grams = moles x molar mass
grams = 0.04025 mol x 119.00 g/mol
grams = 4.79175 g

Rounding off, the answer is 4.79 g, which is choice C.

How Many Grams of KBr Are Required to Make 350