A student dissolves 18.9 grams of Zn(NO3)2 [molar mass = 189] in enough water to make 500.0 of solution. A 100.0 ml portion of this solution is diluted to 1000.0 ml.What is the [NO3¯ ] in the second solution?

Question

A) 0.0100M B) 0.0200M C) 0.0300M
D) 0.0400M E) 0.0500M

Answer 1

Convert 18.9 g Zn(NO3)2 to moles. moles = grams/molar mass

Nitrate ion will be twice that since there are two moles NO3^- to each mole of Zn(NO3)2.

Placing that in 500 mL will be
M = moles/0.500 L = ?? M.

You then dilute 100 mL of that soln to 1000 mL; therefore, the new concn will be ??M x (100/1000) = zz M.