Answer to Question #190071 in General Chemistry for Elisa Giuliani

Question

What mass of silver nitrate AgNo3, is needed to make 0.50L of a 0.10mol/L solution?

Answer 1

Let $x$ be number of moles in 0.50L of solution , to find the value of $x$ molarity of both the solution will be same $\ 0.10=\dfrac{x}{0.50}$ ,$\ x=0.05\ mol$

Now you can calculate mass of silver nitrate , n = $\dfrac{m}{Mw}$ , Where n = number of moles

m = mass

Mw = molecular weight = Molecular weight of $AgNO_3=169.87\ MW$

$m=n\times Mw=$ 169.87$\times0.05=$ 8.4935 g