Question

How many grams of AgNO3 are needed to prepare a 0.25 m solution in 500 grams of water? 125 g 0.125g 21.25 g 170g

Answer 1

Well...`"molality"="moles of solute"/"kilograms of solvent"`

Explanation:

And we require a molar quantity of silver nitrate sufficient to make a `0.25*mol*kg^-1` solution with `500*g` water solvent...

And thus `0.250*mol*kg^-1=(("mass of silver nitrate")/(169.87*g*mol^-1))/(1/2*kg)`..

And so....

`0.250*mol*kg^-1xx1/2*kgxx169.87*g*mol^-1="mass"_(AgNO_3)`

`=21.23*g`...so I think the choice of your options is fairly clear...