Answer to Question #170742 in General Chemistry for Leslie

Question

If you are given 237.74 g of the compound Lead (II) nitrate, keep answers to two decimal places.

How many moles would that contain?

How many particles would be present?

How many nitrate ions would be present?

Answer 1

Mole = mass / molar mass =237.74/331g = 0.72 mole

No of particles = number of mole × avogadro constant

= 0.72 × 6.02×10²³ particles = 4.33 × 10²³ particles

There are 2 moles of nitrate ions in 1 mole of Pb(NO3)2

Therefore, = 2 × 4.33 ×10²³ Ions = 8.66 ×10²³ ions