Question

Suppose you wanted to find out how many milliliters of 1.0 m agno3 are needed to provide 169.9 g of pure agno3? what is step 1 in solving the problem? calculate moles agno3 needed what is the molar mass of agno3? 169.87 g/mol how many milliliters of solution are needed?

Answer 1

m(AgNO₃) = 169.9 g; mass.
n(AgNO₃) = m(AgNO₃) ÷ M(AgNO₃).
n(AgNO₃) = 169.9 g ÷ 169.87 g/mol.
n(AgNO₃) = 1 mol; amount of silver nitrate.
c(AgNO₃) = 1 M = 1 mol/L.
V(AgNO₃) = n(AgNO₃) ÷ c(AgNO₃).
V(AgNO₃) = 1 mol ÷ 1 mol/L.
V(AgNO₃) = 1 L.
V(AgNO₃) = 1 L · 1000 mL/L.
V(AgNO₃) = 1000 mL; volume of silver nitrate.