What volume of 0.450 M barium nitrate solution is needed to prepare 261.0 mL of a solution that is 0.272 M in nitrate

78.8 mL .

Explanation:

Ba( NO₃ )₂ ⇄ Ba⁺² + 2 NO₃⁻

1 mole                        2 mole

261 mL of .272 M in nitrate contains .261 x .272 gram -ions ( moles ) of NO₃⁻

= .071 gram-ions .

Let the volume of Ba(NO₃)₂ required be v litre .

moles of Ba( NO₃ )₂ in v volume = v x .45 moles

v x .45 moles of Ba( NO₃ )₂ will give 2 x  v x .45 moles of NO₃⁻

According to question

2 x  v x .45 =  .071

v = .07888 litre

= 78.8 mL .