What is the molarity of NO 3 – in each solution? a. 0.225 M NaNO 3 b. 0.225 M Mg(NO 3 ) 2 c. 0.225 M Al(NO 3 ) 3

Alright. Hi, everyone. So this question is asking us, what is the molarity of nitrate? That's no three negative in each solution. For part one, we have 0.250 molar of sodium nitrate. For part two, we have 0.250 molar of barium nitrate. And for part three, we have 0.250 molar of aluminum nitrate. And on the screen right now are four different answer choices labeled A through D. Now for a question like this one where our task is to find the concentration of a particular ion. We're really looking for the osmolarity because recall that the osmolarity is equal to the number of moles of a given ion per every liter of solution. So because we're given the molarity for three different solutions, we're going to multiply the number of ions by the molarity of each solution starting off with the first solution which is sodium nitrate. Now, when sodium nitrate dissociates, it dissociates into one sodium cion and one nitrate anion. So to find the osmolarity of nitrate, specifically, we're going to multiply the molarity of the solution that's 0.250 moller by the number of nitrate ions in the solution itself. So thats one and so this equals 0.250 molar or nitrate specifically. So now let's do the same thing for part two, which is barium nitrate. Now, barium nitrate dissociates to form one barium cion with a charge of positive two and two nitrate anions with a charge of negative one each. So for the osmolarity, the molarity of the solution for part two was once again, 0.250 molar. But this time we have two ions of nitrate. So the molarity gets multiplied by two and that equals 0.500 molar nitrate. And so for part three, we have aluminum nitrate. So aluminum nitrate dissociates to form one aluminum cion with a charge of positive three and three nitrate anions. So for the osmolarity were going to take the molar concentration of the solution that's 0.250 molder and multiply this by three ions of nitrate and this equals 0.750 molar of nitrate and there you have it. So here our answer is going to be option B in the multiple choice because for part one, we had 0.250 molar of nitrate. Part two was 0.500 molar. In part three was 0.750 molar. So with that being said, thank you so very much for watching and I hope you found this helpful.

Ch.5 - Introduction to Solutions and Aqueous Solutions

Tro - Chemistry: A Molecular Approach 6th Edition

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Ch.5 - Introduction to Solutions and Aqueous Solutions

Problem 25

Chapter 5, Problem 25

What is the molarity of NO₃^– in each solution? a. 0.225 M NaNO₃ b. 0.225 M Mg(NO₃)₂ c. 0.225 M Al(NO₃)₃

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Identify the number of nitrate ions (NO_3^–) produced per formula unit of each compound.

For NaNO_3, note that it dissociates into one Na^+ ion and one NO_3^– ion.

For Mg(NO_3)_2, note that it dissociates into one Mg^2+ ion and two NO_3^– ions.

For Al(NO_3)_3, note that it dissociates into one Al^3+ ion and three NO_3^– ions.

Calculate the molarity of NO_3^– by multiplying the given molarity of each compound by the number of NO_3^– ions produced per formula unit.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (M) and is crucial for understanding how much of a substance is present in a given volume of solution. In this question, calculating the molarity of NO3– requires knowing the total moles of nitrate ions contributed by each compound.

Dissociation of Ionic Compounds

Ionic compounds dissociate into their constituent ions when dissolved in water. For example, NaNO3 dissociates into Na+ and NO3–, while Mg(NO3)2 dissociates into Mg2+ and two NO3– ions. Understanding how many nitrate ions are produced from each compound is essential for determining the molarity of NO3– in the solutions provided.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. In this context, it helps in determining the number of moles of NO3– produced from each compound based on their dissociation. This is vital for accurately calculating the molarity of the nitrate ion in each solution.