Thursday

May 26, 2016
Posted by **James** on Sunday, March 17, 2013 at 1:46pm.

The question is...

"One troy ounce of gold is worth $380. There are 33.8 grams per troy ounce. You have a stock colloidal solution of a very fine gold dust. It's concentration is .10% w/v. From this solution 1.0 ml is placed in 9.0 ml of an isotonic, pH buffered solution to make a working stock. An assay calls for .50 ml of this working stock to which is added .10 ml of serum. What is the cost of gold used for each test?

My work:

.10 % w/v gold = .10 g gold (in 1 ml)

(.10 g gold)/(1 ml + 9 ml) = .01% w/v gold (or .01 g gold)

(.01 g gold)/(1 ml) x (5 ml)/(1) = .005 g gold in final solution

.005 g gold x (1 troy ounce)/(33.8 g gold) = .0001479 troy ounces of gold in the final solution

.0001479 x $380 = about $.06 per test

Any guidance would be greatly appreciated. Thanks!

- College Chemistry/Lab Math -
**James**, Sunday, March 17, 2013 at 1:48pmSorry, in my 3rd step it should be "(.5ml)/(1)"

- College Chemistry/Lab Math -
**Devron**, Sunday, March 17, 2013 at 3:12pmThis is good, but mass/volume %= (g/mL)*100

So,

(0.10 % w/v)/100 = 0.001 g gold (in 1 ml)

And so,

(0.001g gold)/(1 ml + 9 ml) = 0.0001g of gold/mL

****Don't worry about mass/volume % at this point.

But now you have a solution that is 0.0001g of gold/mL

You remove 0.5mL of solution and add it to 0.10 mL of serum.

How much gold is in a total volume of 0.60 mL?

0.0001g of gold/mL *(0.10mL)=0.00001g of gold

Your new concentration will be

0.00001g of gold/total volume=0.00001g of gold/0.60mL=1.67 x 10^-5 g of gold/mL

or 1.67 x 10^-5 g of gold/mL*(100)=1.67 x 10^-3 % m/v

So, each test uses 0.00001g of gold

0.00001g of gold*(1 troy ounce)/(33.8 g gold) = 2.96 x 10^-7 troy ounces of gold in the final solution

(2.96 x 10^-7)* ($380/oz) = $1.12 x 10^-4

The test isn't that expensive. - College Chemistry/Lab Math -
**James**, Sunday, March 17, 2013 at 6:35pmWhy did you multiply 0.0001g of gold/mL by 0.10mL? Shouldnt it be .50 mL? I say this because .10 is the volume of serum added, which does not contribute to the total amount of gold in the final solution.

- College Chemistry/Lab Math -
**Devron**, Sunday, March 17, 2013 at 7:18pmThank you for catching that.

0.0001g of gold/mL *(0.50mL)=0.00005g of gold

Your new concentration will be

0.00005g of gold/total volume=0.00005g of gold/0.60mL=8.33 x 10^-5 g of gold/mL

or 8.33 x 10^-5 g of gold/mL*(100)=8.33 x 10^-3 % m/v

So, each test uses 0.00005g of gold

0.00005g of gold*(1 troy ounce)/(33.8 g gold) = 1.48 x 10^-5 troy ounces of gold in the final solution

(1.48 x 10^-5)* ($380/oz) = $5.62 x 10^-3 - College Chemistry/Lab Math -
**James**, Monday, March 18, 2013 at 9:18amThanks for the help!

- College Chemistry/Lab Math -
**Kip**, Saturday, April 6, 2013 at 12:54pmgo back to basics and CHECK YOUR NUMBERS. There are only 31.1 grams in a troy ounce my friend...