Question

How many moles of nitrate ions are present in exactly 275 ml of a 1.25 m copper(ii) nitrate solution, cu(no3)2 (aq)?

Answer 1

To find the number of moles of nitrate ions in 275 mL of a 1.25 M copper(II) nitrate solution, you multiply the volume of the solution (converted to liters) by the molarity, and then by 2, as each formula unit of copper(II) nitrate produces two nitrate ions. The result is 0.6875 moles of nitrate ions.

Step-by-step explanation:

To determine how many moles of nitrate ions are present in exactly 275 mL of a 1.25 M copper(II) nitrate solution, Cu(NO3)2 (aq), you first need to calculate the moles of copper(II) nitrate in the solution. Since the solution is 1.25 M, this means that for every liter of solution, there are 1.25 moles of copper(II) nitrate. Since there are 1000 mL in a liter, we can convert milliliters to liters by dividing by 1000: 275 mL = 0.275 L

Multiplying the volume of the solution in liters (0.275 L) by the molarity (1.25 M), we will find the moles of Cu(NO3)2:

0.275 L × 1.25 M = 0.34375 moles of Cu(NO3)2

As copper(II) nitrate dissociates into one Cu2+ ion and two NO3− ions, we need to multiply the moles of Cu(NO3)2 by 2 to find the moles of nitrate ions:

0.34375 moles of Cu(NO3)2 × 2 = 0.6875 moles of NO3− ions

Answer 2

The formula for copper nitrate is written as Cu(NO₃)₂. The formula for nitrate is NO₃. So, that means that for every 1 mole of Cu(NO₃)₂, there are 2 moles of NO₃. The ratio is 1:2. Also, the concentration in molarity is expressed as moles solute per liter solution. The solution is as follows:

Moles nitrate = 1.25 mol Cu(NO₃)₂/L * 1 L/1000 mL * 275 mL * 2 mol NO₃/1mol Cu(NO₃)₂ = 0.6875 moles