Given:
Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL
The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:
45ml * 0.10 M = C analyte * 20 ml
C analyte = 0.225 M
Answer:
m= 4,599.145 g
Explanation:
Let m = mass, d = density and V = volume of the osmium block.
m = d x V
m = 22.610 g/cm3 x (6.70 x 9.20 x 3.3) cm3
m = 4,599.145 g
