Unit Cell Volume

The smallest repeating unit of s crystal of table salt(the unit cell) is a cube with an edge of .0563nm. If the density of NaCl is 2.17 g/cm^3, what is the mass in grams of one unit cell?
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The edge of a unit cell in meters is:
1 nm = 1.00x10^-9m
Since 1 m = 100 cm = 10^2 cm,
(1.00x10^-9m)(10^2cm/m) = 1.000x10^-7 cm
The volume of a unit cell is:
(1.00x10^-7cm)^3 = 1.00x10^-21 cm^3

mass(unit cell) = (2.17 g/cm^3)(1.00x10^-21 cm^3) = 2.17x10^-21 g

Saline Solution

On 2/4/08 2:30 PM, "emaildave" wrote:

Physiological saline is a .9% solution of NaCl. (a) Express this in terms of molar or millimolar concentrations. (b) Make 125 mL of physiological saline.

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(a) 0.9% means 0.90g NaCl / 100 mLs

The above concentration is equivalent to 9.0g / 1000mLs or 9.0g / liter

The molar mass of NaCl is 23.0 + 35.45 = 58.45 g/mole of NaCl

(9.0g/L) / (58.45 g/mol) = 0.154 moles NaCl / liter or 154 millimoles/liter

NOTE:
1 Mole of NaCl contains 2 moles of ions (Na+ and Cl-). The above molar concentration is equivalent to:
(0.154 mol/L)(2 moles of ions/mol) = 0.308 moles of ions / liter.
This adjusted molarity is called “osmolarity”. “Colligative” properties such as osmotic pressure, freezing point depression, boiling point elevation, vapor pressure, etc. depend on osmolarity rather than molarity.

(b) (0.125L)(0.154mol/L)((58.45 g/mol) = 1.125g/L NaCl

pH During Titration

Consider the titration of 100 mL of .500 M NH3 (kb=1.8x10^-5) with .500 M HCl. After 50.0 ml of HCL have been added, the [H+] of the solution is a) 1.8x10^-5 b) 5.6x10^-10 c) 1.2x10^-5 d) 1.0x10^-7 e) none of these
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NH_3(aq) + H₂O <==> NH₄⁺(aq) + OH-(aq)
K_b = [ NH4+][ OH-]/[ NH3] = 1.8x10^-5
(0.100L)(0.50mol/L) = 0.050 moles NH3
(0.050L)(0.50mol/L) = 0.025 moles HCl
0.025 moles of NH3 will be converted to NH4+ when NH3 reacts with HCl
We will end up 0.025 moles of each, NH3 and H+, in 150 mLs total volume
The concentrations are:
[NH3] = [H+] = 0.025 moles / 0.150L = 0.167 M
[ 0.167][ OH-]/[0.167] = 1.8x10^-5, SOLVE for [OH-]
[OH-] = 1.8x10^-5
[H+][1.8x10^-5]=1x10^-14
[H+]=5.55x10^-10

Dilution Calculations

You only have 2mL of a 1M NaCl stock solution. You need 100mL of a .005 M NaCl solution. Can it be done with the volume of stock solution you have? Show your calculations then do the dilution.

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1. A helpful formula for this type of calculation is:
V1M1 = V2M2
Substituting,
(2 mL)(1M) = (V2)(0.005)
V2 = 400 mLs (that we can prepare, more than enough for 100mls).

If we just want to prepare 100 mLs.,
(V1)((1M) = (100mLs)(0.005M)
V1 = (100mLs)(0.005M) / (1M) = 0.5 mL needed

Average Isotopic Mass

Silver has two table isotopes. The actual isotopic mass of ¹⁰⁷Ag is 106.9051 amu and the isotopic mass of ¹⁰⁹Ag, 108.9047amu. What is the percent abundance of each isotope?
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The average atomic mass of Ag is 107.8682 (you can look this up).
Let the fraction Ag-109 = x (this is really the % expressed as a decimal fraction)
Then, the fraction of Ag-107 would be 1-x (1 means 100%)
Now set up:
108.9047x + 106.9051(1-x) = 107.8682
(This is an expression for the “weighted average” which is given so we calculate the percent of each isotope)

We simplify the above equation and solve for x,
then we substitute below to get the percents
100x = % of Ag-109
(100)(1-x) = % of Ag-107