Question

What is the nitrate ion concentration, [NO3¹], in 100.mL of magnesium nitrate

solution, Mg(NO3)2, that has 2.5g dissolved magnesium nitrate?

Answer 1

The nitrate ion concentration in 100 mL of magnesium nitrate solution, Mg(NO3)2, that has 2.5g of dissolved magnesium nitrate is 0.169 M.

Step-by-step explanation:

To determine the nitrate ion concentration, [NO3¹], we need to use the information provided about the magnesium nitrate solution, Mg(NO3)2. The molar mass of Mg(NO3)2 is 148.32 g/mol. We are given the mass of the dissolved magnesium nitrate as 2.5 g and the volume of the solution as 100 mL.

To find the concentration, we can first convert the mass of the dissolved magnesium nitrate to moles using the formula:

Moles = Mass / Molar Mass

Converting the mass to moles gives us:

Moles = 2.5 g / 148.32 g/mol = 0.0169 mol

Next, we can use the volume of the solution to calculate the concentration:

Concentration = Moles / Volume = 0.0169 mol / 0.1 L (since 100 mL = 0.1 L) = 0.169 M

Therefore, the nitrate ion concentration in 100 mL of magnesium nitrate solution is 0.169 M.

Answer 2

Concentration = 0.0337 mol / 0.1 L = 0.337 M

Therefore, the nitrate ion concentration, [NO3¹], in 100 mL of magnesium nitrate solution that has 2.5 g dissolved magnesium nitrate is approximately 0.337 M.

Step-by-step explanation:

To find the nitrate ion concentration, [NO3¹], in the magnesium nitrate solution, we need to use the given information and perform a few calculations.

First, we need to determine the number of moles of magnesium nitrate present in the solution. We can do this by dividing the mass of magnesium nitrate by its molar mass. The molar mass of magnesium nitrate (Mg(NO3)2) can be calculated by adding up the atomic masses of its elements:

Mg(NO3)2 = (1 x Mg) + (2 x N) + (6 x O) = 24.31 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol) = 148.33 g/mol

Next, we can calculate the number of moles of magnesium nitrate by dividing the given mass (2.5 g) by its molar mass:

Number of moles = Mass / Molar mass = 2.5 g / 148.33 g/mol ≈ 0.01685 mol

Since 1 mole of magnesium nitrate (Mg(NO3)2) contains 2 moles of nitrate ions (NO3¹), the number of moles of nitrate ions is twice the number of moles of magnesium nitrate:

Number of moles of NO3¹ = 2 x Number of moles of Mg(NO3)2 = 2 x 0.01685 mol ≈ 0.0337 mol

Now, we need to calculate the concentration of nitrate ions ([NO3¹]) in the solution. Concentration is defined as the number of moles of solute divided by the volume of the solution. In this case, we have 0.0337 moles of NO3¹ in a 100 mL solution:

Concentration = Number of moles / Volume

Concentration = 0.0337 mol / 100 mL

However, it is best to express the volume in liters, so we convert 100 mL to 0.1 L:

Concentration = 0.0337 mol / 0.1 L = 0.337 M