Answer:
Explanation:
Solution required is .15 M 500 mL sodium fluoride .
molecular weight of sodium fluoride = 42
one mole of sodium fluoride = 42 grams
No of moles contained in the solution of .15M , 500 mL solution
= .15 x 0.5 moles = .075 moles
1 mole = 42 g
.075 moles = .075 x 42 = 3.15 grams .
So mass of sodium fluoride required = 3.15 g .
Thietbi od
0.0294
Mass= 2.77g
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g