Chemistry Help Forum Solubility and concentration

 Sep 29th 2017, 09:31 PM #1 PrincessPeach Junior Member   Join Date: Sep 2017 Location: Earth Posts: 2 Solubility and concentration Hi I am confused on what formulas to use when answering questions like these: 1. A sample of pond water contained 88.00 µg g−1 of dissolved oxygen gas. Calculate the amount in grams of dissolved oxygen in 4.00 L of pond water. 2. A desirable level of dissolved oxygen in river water is 8 ppm. A sample of river water is found to have less than this, with 5.60 ppm oxygen. If 1.40 µg of dissolved oxygen was actually present in the sample, calculate the mass of water tested. If someone could help me or give me some steps to solve the problems that would be great. Please explain simply
Sep 29th 2017, 10:40 PM   #2
oz93666
Member

Join Date: May 2017
Location: thailand
Posts: 75
 Originally Posted by PrincessPeach Hi 1. A sample of pond water contained 88.00 µg g−1 of dissolved oxygen gas.
So that's 88 micrograms per gram ...( i'm not familiar with that notation g-1 , I'll read it as "per gram" )

A microgram is a millionth of a gram ... 0.000088 gram per gram of water

or 0.088grams per Kg of water

pond water has a density of about 1 , so 1Liter is 1 Kg

But we have 4 liters of water ....4x 0.088 =0.352grams is the answer

Sep 29th 2017, 10:53 PM   #3
oz93666
Member

Join Date: May 2017
Location: thailand
Posts: 75
 Originally Posted by PrincessPeach 2. A desirable level of dissolved oxygen in river water is 8 ppm. A sample of river water is found to have less than this, with 5.60 ppm oxygen. If 1.40 µg of dissolved oxygen was actually present in the sample, calculate the mass of water tested.
a microgram is a millionth of a gram so 1ppm is the same as 1 ug per gram

We are told the oxygen level is 5.6 ppm ... every gram of water has 5.6ug

but only 1.4 ug of oxygen ... how much water will hold this amount ...

1.4 divided by 5.6...= 0.25 grams

 Thread Tools Display Modes Linear Mode