Question

13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solution?

A. 0.00741 M
B. 0.119 M
C. 0.135 M
D. 24.6 M

Can you explain how you would set up the equation?

Answer 1

Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate
`Pb(NO_3)_4`
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of
`Pb(NO_3)_4` = 207+56+192 = 455 g/mol

Formula:

`M = (m)/(MM*V)`

Solving:

`M = (m)/(MM*V)`

`M = (13.50)/(455*0.25)`

`M = (13.50)/(113.75)`

`M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark`

Answer:
B. 0.119 M