If a chemist has a stock solution of hbr that is 10.0 m and would like to make 450.0 ml of 3.0 m hbr, how would he do it?

This is a dilution that requires a certain volume from the stock solution to be diluted with distilled water to make a solution of HBr with a lesser concentration than the stock solution

Following dilution formula can be used

c1v1 = c2v2

Where c1 is concentration and v1 is the volume of the stock solution

c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting these values

10.0 M x v1 = 3.0 x 450.0 mL

v1 = 135.0 mL

A volume of 135.0 mL from HBr stock solution needs to be taken and diluted with distilled water upto 450.0 mL. The resulting solution will have a concentration of 3.0 M

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.