Which barium nitrate solution requires the greatest moles of solute?

a. 2.58 l of 0.0250 m ba(no3)2

b. 19.23 ml of 8.5 × 10−2 m ba(no3)2

c. no right answer.

d. 26.20 ml of 2.21 m ba(no3)2

e. 1.77 l of 1.55 m ba(no3)2

The answer to your question is: letter E,  1.77 L of 1.55 M Ba(NO3)2

Explanation:

Formula

Molarity =

moles = Molarity x volume (L)

a. 2.58 L of 0.0250 M Ba(NO3)2

moles = (0.025) (2.58)

moles = 0.065 M

b. 19.23 mL of 8.5 × 10−2 M Ba(NO3)2

moles = (8.5 x 10 ⁻²) (0.01923)

moles = 0.0016

c. No right answer.

d. 26.20 mL of 2.21 M Ba(NO3)2

 moles = (2.21)(0.0262)

moles = 0.058

e. 1.77 L of 1.55 M Ba(NO3)2

  moles = (1.55)(1.77)

  moles = 2.74            This solution needs the greatest concentration of

                                    Ba(NO₃)₂

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