Answer:
( About ) 5.7 grams
Explanation:
[tex]Molarity = 0.135 mol L^{-1}[/tex]
Every [tex]1L = 10^3 ML[/tex]
.135 Moles of Agno3
[tex]250 ML = \frac{10^3ML}{4}[/tex]
[tex]\frac{0.135}{4} = (0.03375 Moles of Agno3)[/tex]
[tex]0.03375 m[/tex] · [tex]\frac{169.87g}{1 M}[/tex]
[tex]0.03375 m[/tex] · [tex]\frac{169.87g}{1 M} = 5.7331125[/tex]
[tex]Solution = About 5.7 Grams[/tex]
*169.87g = weight of silver Nitrate (Agno3)
