Answer:
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 .
