What is the molarity of nitrate ions in a solution containing 2.20 mg of Mg(NO3)2 dissolved in 7.35 L of water?
2.20 x 10^-3g *(1mole/148.3 g of Mg(NO3)2)= moles of Mg(NO3)2
moles of Mg(NO3)2 *(2 moles of NO3/1 mole of Mg(NO3)2))= moles of NO3
moles of NO3/7.35L=molarity
Thank you so much!
To determine the molarity of nitrate ions, we need to first calculate the number of moles of Mg(NO3)2 in the given solution, and then divide it by the volume of the solution.
Step 1: Convert the mass of Mg(NO3)2 to moles.
The molar mass of Mg(NO3)2 can be calculated by adding the atomic masses of each element.
Mg(NO3)2 = Mg + 2(NO3)
Molar mass of Mg(NO3)2 = (24.31 g/mol) + 2 * [(14.01 g/mol) + (3 * 16.00 g/mol)]
Molar mass of Mg(NO3)2 = 24.31 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
Molar mass of Mg(NO3)2 = 24.31 g/mol + 2 * 62.01 g/mol
Molar mass of Mg(NO3)2 = 24.31 g/mol + 124.02 g/mol
Molar mass of Mg(NO3)2 = 148.33 g/mol
Now we can use the molar mass to convert the given mass of Mg(NO3)2 to moles:
Mass of Mg(NO3)2 = 2.20 mg = 0.00220 g (since 1 mg = 0.001 g)
Number of moles of Mg(NO3)2 = 0.00220 g / (148.33 g/mol)
Number of moles of Mg(NO3)2 = 0.0000148 mol
Step 2: Calculate the molarity of nitrate ions.
Molarity is defined as moles of solute divided by volume of solution in liters.
Molarity (M) = Number of moles of solute / Volume of solution (in liters)
Volume of solution = 7.35 L
Molarity of nitrate ions = Number of moles of Mg(NO3)2 / Volume of solution
Molarity of nitrate ions = 0.0000148 mol / 7.35 L
Molarity of nitrate ions = 2.02 x 10^-6 M
Therefore, the molarity of nitrate ions in the solution is 2.02 x 10^-6 M.
To calculate the molarity of nitrate ions in the solution, we need to know the molar mass of magnesium nitrate, Mg(NO3)2. Let's break down the calculation step by step:
Step 1: Convert the given mass of Mg(NO3)2 to moles.
To do this, we need to divide the mass by the molar mass of Mg(NO3)2. The molar mass of magnesium nitrate is the sum of the atomic masses of each element in the compound.
Molar mass of Mg(NO3)2 = (24.31 g/mol for Mg) + (14.01 g/mol for N) + (3 * 16.00 g/mol for O)
= 24.31 g/mol + 14.01 g/mol + 48.00 g/mol
= 86.32 g/mol
Now, let's convert the given mass of Mg(NO3)2 to moles:
Moles of Mg(NO3)2 = (2.20 mg) / (86.32 g/mol) * (1 g / 1000 mg)
= 0.02547 mol
Step 2: Calculate the molarity of the solution.
Molarity (M) is defined as the number of moles of solute per liter of solution. In this case, the solute is nitrate ions (NO3-), and the solution volume is given as 7.35 L.
Molarity of nitrate ions = (moles of NO3-) / (volume of solution in liters)
Since magnesium nitrate dissociates into two nitrate ions (NO3-) per formula unit, we need to multiply the moles of Mg(NO3)2 by 2 to get moles of nitrate ions.
Moles of NO3- = 2 * 0.02547 mol
= 0.05094 mol
Molarity of nitrate ions = (0.05094 mol) / (7.35 L)
= 0.00694 M
So, the molarity of nitrate ions in the solution is 0.00694 M.