Respuesta :
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
Answer:
V_{solution}=0.355L
Explanation:
Hello,
Potassium bromide moles are computed as:
[tex]n_{KBr}=30.5gKBr*\frac{1molKBr}{119.9gKBr}=0.254gKBr[/tex]
Now, by recalling the molarity formula and solving for the volume, one obtains:
[tex]M=\frac{n_{KBr}}{V_{solution}} \\V_{solution}=\frac{n_{KBr}}{M}\\V_{solution}=\frac{0.254mol}{0.716mol/L}\\ V_{solution}=0.355L[/tex]
Best regards.