question below with answers

Question

What volume (in milliliters) of 2.50 M lead (II) nitrate solution contains 0.0500 moles of Pb²⁺ ions?

1. a. .02 mL
2. b. .04 mL
3. c. 20 mL
4. d. 40 mL

Answer 1

Lead(ll) nitrate = Pb(NO₃)₂.

This dissociates as follows: Pb(NO₃)₂ => Pb²⁺ + 2NO₃^-

2.5 M = 2.5 moles/liter

(x L)(2.5 mol/L) = 0.0500 moles Pb²⁺

x = 0.02 liters

0.02 L x 1000 ml/L = 20 mls <—ANSWER c