   # What is the mass of solute, in grams, in 125 mL of a 1.023 × 10 –3 M solution of Na 3 PO 4 ? What is the molar concentration of the Na + and PO 4 3– ion? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 42PS
Textbook Problem
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## What is the mass of solute, in grams, in 125 mL of a 1.023 × 10–3 M solution of Na3PO4? What is the molar concentration of the Na+ and PO43– ion?

Interpretation Introduction

Interpretation:

The mass of solute in grams, in 125mL of a 1.023×10-3M solution of Na3PO4 and the molar concentration of Na+andPO43- ions should be determined.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one litre of the solution.

Molarity=MassperlitreMolecular mass

• A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,

Numberofmole=GivenmassofthesubstanceMolarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

### Explanation of Solution

Molar concentration of sodium ion [Na+]= 3×0.0012MNa+(aq)=0.0036M

Molar concentration of phosphate ion [PO43-] =0.0012M

Molar concentration of given Na3PO4 solution is 0.001023M and the volume of the solution in litre is .125L.

Before to find the mass of solute in gram, the amount of Na3PO4 required to make 0.001023M solution has to be determined.

Thus,

Amount of Na3PO4 required = .125L×0.001023molofNa3PO41Lsolution=0.0001278mol

The mass of solute (Na3PO4) in grams can be calculated as follows,

0.0001278molNa3PO4×163.94gNa3PO41molNa3PO4=0.021gNa3PO4

Therefore the mass of solute, in grams, in .125L of a 0.001023M solution of Na3PO4 is 0.021g.

The molar concentrations of Na+andPO43- is,

Amount of Na3PO4 (in mol) can be calculated as follows,

0

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