111k views
3 votes
How many mol of lead nitrate are present in 12.29 mL of a 0.436 M solution?

User J Pullar
by
5.0k points

1 Answer

1 vote

Answer:

0.00536 moles Pb(NO₃)₂

Step-by-step explanation:

To find the amount of moles, you need to (1) convert milliliters (mL) to liters (L) and then (2) calculate the number of moles (using the molarity ratio). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs.

(Step 1)

1,000 mL = 1 L

12.29 mL 1 L
----------------- x ------------------- = 0.01229 L
1,000 mL

(Step 2)

Molarity = moles / volume (L) <----- Molarity ratio

0.436 M = moles / 0.01229 L <----- Insert given values

0.00536 = moles <----- Multiply both sides by 0.01229 L

User Henning Koehler
by
4.6k points