A 30.2 mL aliquot of a 30.0 wt% aqueous KOH solution is diluted to 1.20 L to produce a 0.173 M KOH solution. Calculate the density of the 30.0 wt% KOH solution.

The density of solution is 1.283 g/mL.

Explanation:

Molarity of the KOH before dilution =

Volume of the solution before dilution =

Molarity of the KOH after dilution =

Volume of the solution after dilution =

(1 mL = 0.001 L)

Mass of 0.2076 moles of KOH:

0.2076 mol × 56 g/mol = 11.6256 g

Mass of KOH is solution = 11.6265 g

Mass of the solution = M

Mass percentage of solution = 30.0% of KOH

M = 38.755 g

Density of the solution , d=

The density of solution is 1.283 g/mL.

you need to give more information

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !