Question

How would you prepare 175.0mL of 0.150M AgNO3 solution starting with pure solute?

Answer 1

@oOKawaiiOo

Answer 2

ok can u make sure from my steps

first i should find the volume of the solution

Answer 3

i will show you let me write it down. hold on

Answer 4

ok

Answer 5

\[1000 ml = 1L\]
\[175.0ml = 0.175\]
\[0.175 L \times \frac{ .150 }{ L } = 0.02625 moles \times \frac{ 170g }{ moles } = 4.4625 grams\] of pure AgNO3