12% NaCl is the answer I came to... Have a GREAT day!!! :)
M1 = 17.45 M
M2 = 0.83 M
V2 = 250 ml
M1. V1= M2. V2
V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml
Answer:
33.95 grams of NaN3
Explanation:
Number of moles of NaN3 = mass (m)/MW = m/65 mole
I mole of NaN3 requires 22.4L air bag
m/65 mole of NaN3 required 11.7L
22.4m/65 = 11.7
22.4m = 65×11.7
22.4m = 760.5
m = 760.5/22.4 = 33.95grams of NaN3