If you do not have the means to weigh the chemical, and if an approximate concentration is needed, than use the approximate value of density=1.840 g/mL to estimate the weight from given volumes. If you need a modified version without weighing, please let us know.
It is ironic that you need w/w concentration but do not have access to mass determination. I hope you do have access to volume measurements!
I have asked this question before but I was not clear enough. I want to prepare a 37% w/w solution (in water)of sulfuric acid from 96% sulfuric acid. I need around 500ml, I know it won't be exactly that, I want to know how to use the conversion using density, I've been shown a way I think is incorrect and need some input. I cannot weigh the chemical I need to have a conversion equation.
2 answers
Here it is:
H2SO4:
500 g 35% w/w H2SO4 contains
500*0.37 g pure H2SO4, or
500*0.37/0.96 g of 96% (w/w) H2SO4
=192.71 g of 96% H2SO4
=192.71/1.840 mL of 96% H2SO4
=104.73 mL. of 96% H2SO4
H2O:
Volume of diluting water
= 500-192.71 g
=307.29 g
= 307.29 mL.
Summary:
use
104.73 mL 96% (w/w) H2SO4
307.29 mL H2O
to make 500 g of 37%(w/w) H2SO4
(all approximate)
H2SO4:
500 g 35% w/w H2SO4 contains
500*0.37 g pure H2SO4, or
500*0.37/0.96 g of 96% (w/w) H2SO4
=192.71 g of 96% H2SO4
=192.71/1.840 mL of 96% H2SO4
=104.73 mL. of 96% H2SO4
H2O:
Volume of diluting water
= 500-192.71 g
=307.29 g
= 307.29 mL.
Summary:
use
104.73 mL 96% (w/w) H2SO4
307.29 mL H2O
to make 500 g of 37%(w/w) H2SO4
(all approximate)
1 answer