How many milliliters of 1.0m agno3 are needes to provide 169.9 g of agno3?

  The number  of  Ml  of  1.0 AgNo3  needed   to  provide   169.9   0f  AgNO3  is   calculated   as  follows

 find  the  moles  of  AgNo3

=  mass/molar  mass  =   169.9/169.87  =   1  mole

volume in Ml  =  moles/ molarity    x1000

=  1/1  x1000=  1000  ml

The molar mass of silver nitrate is 169.87 g/mol
Therefore the number of moles in 169.9 g will be;
 = 169.9/169.87 
 = 1 mole
Therefore; since the molarity is 1.0 M
then the volume= 1/1 =1 liter which is equivalent to 1000ml
Thus, the answer is 1000 ml

Tolerance to high levels of salinity