My Notes Ask Your A solution is prepared by dissolving 26.8 g of ammonium sulfat

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My Notes Ask Your A solution is prepared by dissolving 26.8 g of ammonium sulfate in enough water to make 100.0 mL of stock solution. A 12.20-mL sample of this stock solution is added to 50.80 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution H4 Need Help? Reatn Supporting Materials Periodic Table constants and Factors My Notes Ash The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every 106 parts of solution. (Both solute and solution are measured using the same units.) Mathematically, by mass, ppm can be expressed as shown below g solution kg solution In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to 1.0,g of solute per 1.0 mL of solution, which equals 1.0 g of solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions (a) 4.2 ppb Hg in H20 (b) 3.0 ppb CHCI3 in H20 (c) 37.0 ppm As in H20 (d) 0.30 ppm DDT (C14HgCls) in H20 Need Help?Re

Explanation / Answer

1.

Given mass of (NH4)2SO4 to moles= 10.81/(132.14)(Molecular weight) ==0.0817 mol (NH4)2SO4=n---(1)

Molarity= Moles of solution/Volume of solution(L)

Given is 100 mL soln, somolarity of the solution

molarity=(n/v)=0.0817/100 soln=0.817M

n=moles,volume

10 mL of this solution is added to 50 mL H2O, which makes a 60-mL total solution.

We can now use the dilution equation

M1V1=M2V2

Molality of the new, 60-mL solution:

(0.0817M)(10)=(M2)(60)

M2=(0.817M)(10mL)60mL=0.136M

This means that there are 0.136 moles of (NH4)2SO4 per liter of solution.

Let's recognize that 1 mol (NH4)2SO4 contains

2 mol NH+4

1 mol SO24

The concentrations of each ion is thus

(2)(0.136M)=0.272M NH+4

(1)(0.136M)=0.136M SO24

2. (a) 4.2 ppb Hg in H2O
Think one liter of the solution, it must contain 4.2x10^-6 g Hg. Since the molar mass of Hg is: 200.59 g/mol, 4.2x10^-6 g Hg is:
(4.2x10^-6 g)/(200.59 g/mol) = 2.1x10^-8 mol.
Since one liter solution contains 2.1x10^-8 mol Hg, the molarity is 2.1x10^-8 M

(b) 3 ppb CHCl3 in H2O
Molar mass of CHCl3 is: 119.38 g/mol
Molarity: 3x10^-6/119.38 = 2.5x10^-8 M

(c) 37ppm in H2O

Molar mass ofH2O is: 18 gm/mol

Molarity= 37x10^-3/18= 2.05 10^-3 M

(d) 0.30 ppm DDT (C14H9Cl5 in H2O)

Molar mass ofH2O is: 18 gm/mol

Molarity=

.3x10^-3/18= 0.0166 10^-3 M

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