What is the concentration (by mass) of the solute in a 25 degree Celsius solution prepared by dissolving 2.50 * 10 ^ -7 mol Na3PO4 in 1.000 L of water (in ppm, ppb, and ppt, all to two decimal places) ? 2.50×10¬ mol Na3PO, in 1.000 L of water (in ppm, ppb, and ppt, all to two decimal places)?A: 4.00 × 10~2 ppm; 4.10 × 10' ppb; 4.10×10* pptWhat is the concentration (by mass) of the solute in a 25 °C solution prepared by dissolvingaosopa to noinlce bota M 0.el cto (Jm m) slov fWTnonios M 2.00ASEra of Lupaadd Water Volume.ConcentrationVolumeA 20.0 mL aliquot of a 0.515 M aqueous solution is added to 125 mL of pure water. What is theconcentration (in M) of the new solution?A: 0.0710Mores of Solure - Concen tration x Vorume0.515M x 20.0mL= 10.3 mmal Volume = (20.0+125 ) mL= 145,0mL%3Dal Concentration = number of moles in So lute%3DTotal Volume10.3 mmolニ146.0mし=6.0710 M

Given : Moles of Na3PO4 dissolved = 2.50 X 10-7 mol. And volume of water = 1.000 L = 1000 mL…

Answered: What is the concentration (by mass) of…

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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What is the concentration (by mass) of the solute in a 25 degree Celsius solution prepared by dissolving 2.50 * 10 ^ -7 mol Na3PO4 in 1.000 L of water (in ppm, ppb, and ppt, all to two decimal places) ?

2.50×10¬ mol Na3PO, in 1.000 L of water (in ppm, ppb, and ppt, all to two decimal places)?
A: 4.00 × 10~2 ppm; 4.10 × 10' ppb; 4.10×10* ppt
What is the concentration (by mass) of the solute in a 25 °C solution prepared by dissolving
aosopa to noinlce bota M 0.el cto (Jm m) slov fW
Tnonios M 2.00ASE
ra of Lupa
add Water Volume.
Concentration
Volume
A 20.0 mL aliquot of a 0.515 M aqueous solution is added to 125 mL of pure water. What is the
concentration (in M) of the new solution?
A: 0.0710
Mores of Solure - Concen tration x Vorume
0.515M x 20.0mL= 10.3 mm
al Volume = (20.0+125 ) mL= 145,0mL
%3D
al Concentration = number of moles in So lute
%3D
Total Volume
10.3 mmol
ニ
146.0mし
=6.0710 M

Expert Solution

Step 1

Given : Moles of Na₃PO₄ dissolved = 2.50 X 10^-7 mol.

And volume of water = 1.000 L = 1000 mL (Since 1 L = 1000 mL)

Since density of water is 1.0 g/mL.

Hence mass of water = density X volume of water = 1.0 X 1000 = 1000 g

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