I think I have this problem figured out correctly, but is there anyone who can tell me if I'm doing something wrong.
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!
6 answers
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.
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