Question

What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?

Answer 1

Answer:

V_{solution}=0.355L

Explanation:

Hello,

Potassium bromide moles are computed as:

$n_{KBr}=30.5gKBr*\frac{1molKBr}{119.9gKBr}=0.254gKBr$

Now, by recalling the molarity formula and solving for the volume, one obtains:

$M=\frac{n_{KBr}}{V_{solution}} \\V_{solution}=\frac{n_{KBr}}{M}\\V_{solution}=\frac{0.254mol}{0.716mol/L}\\ V_{solution}=0.355L$

Best regards.

Answer 2

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.