Question

How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?

Answer 1

0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 × 0.35) = 04025 moles
From the formula of molarity moles = molarity × volume in liters
1 mole of KBr is equivalent to 119 g
Therefore, the mass = 0.04025 ×  119  g = 4.79 g

Answer 2

Answer is: 4.79 grams og KBr are required.

V(KBr) = 350 mL ÷ 1000 mL/L.

V(KBr) = 0.350 L; volume of potassium bromide solution.

c(KBr) = 0.115 M; molarity of potassium bromide solution.

n(KBr) = c(KBr) · V(KBr).

n(KBr) = 0.115 mol/L · 0.350 L.

n(KBr) = 0.04025 mol, amount of substance.

m(KBr) = n(KBr) · M(KBr).

m(KBr) = 0.04025 mol · 119 g/mol.

m(KBr) = 4.79 g; mass of potassium bromide.

M - molar mass.