posted by Anonymous on .
prepare 1M 100 ml H2SO4. 98% H2SO4, density 1.84
First of all, the density makes no sense without units. Secondly, you should only specify the amount of something present in one way. The questions gives two: (1) a molarity and a volume. (2) a percentage (by volume or by mass?) and a density. So you can't solve such a question unambiguously in general.
Suppose we go with the given molarity. That means there are 0.1 moles of sulfuric acid present (1.0 mol/L * 0.1 L). That will require adding how many grams to 100 mL of water? Molecular mass of H2SO4 = 98 g/mol, so use 98*0.1 = 0.98 grams of H2SO4.
If we go with the percent by mass, then assume the density was supposed to be 1.84 g/L, then that's 1.84g/L*0.1L = 0.184 grams all told, 98% of which is H2SO4 = 0.18032 grams.
So you see, the desired solution is impossible to obtain on the face of it. You can either have a desired molarity or a desired mass percentage and density, but not both (unless by luck the desired quantities coincide).
First of all density with no units is meaningless and makes the question impossible to solve.
Secondly the percentage without the type is meaningless.
Although can easily be find out in such question but, you should be specify that.
But to give idea about the solution lets assume
d(H2SO4) = 1.84 kg/L and 98% means
100 unit volume of stock solution contains
98 unit volume of H2SO4
so problem asks us to find the amount (mL) of H2SO4 (stock) to prepare 100 mL of 1M soluiton
x mL stock(H2SO4) =
(1mol H2SO4 / 1L solution) *
(1L solution / 1000mL solution) * (100mL solution) *
(98.04g H2SO4 / 1mol H2SO4) *
(1kg H2SO4 / 1000 g H2SO4) *
(1L H2SO4 / 1.84kg H2SO4) *
(1000mL H2SO4 / 1L H2SO4) *
(100ml stockH2SO4 / 98mL H2SO4)
x mL H2SO4 = 5.437 mL
to prepare 100 mL, 1M H2SO4 from stock (98% [v/v]; 1.84 [kg/L]; 98.04 [g/mol])
to 100 mL flat bottom flask add some amount of water (not much),
and pour 5.45 mL H2SO4 stock solution on it.
Dilute the solution to 100 mL