What mass (in g) of KIO3 is needed to prepare 50.0 mL of 0.20 M KIO3? b) What volume (in mL) of 0.15 m H2SO4 is needed to prepare 50.0 mL of 0.080 M H2SO4?

a) mass = 2.14 g

b) volume = 14.76 mL

Explanation:

a) mass of KIO3:

In order to know the mass of any compound, we need the molar mass and the moles so:

m = n * MM

to calculate the moles, we already have the concentration of the solution required and it's volume, so, from there, we can calculate the moles:

n = M * V

Replacing the data (And converting mL to L, dividing by 1000):

n = 0.2 * 0.050 = 0.01 moles

Now, the reported molar mass of KIO3 is 214 g/mol so, the mass:

m = 0.01 * 214

m = 2.14 g of KIO3 are needed to prepare this solution

b) volume of H2SO4 needed:

In this case, we have a little issue, the concentration of the H2SO4 given is expressed in molality (m) and not molarity (Capital M), so, in order to convert this to molarity, we need the density. The density of H2SO4 we need to use here is 1.83 g/mL, so, let's convert the molality to molarity and then, the volume:

m = n/kg solvent

Now, 0.15 m we can rewrite this like:

0.15 moles solute / kg solvent

In this case, we have 0.15 moles in 1 kg of solvent or 1000 g (Which we can assume is water).

Now, that we have the moles, let's calculate the mass of acid, with the molecular weight (MM = 98 g/mol)

mass = 0.15 * 98 = 14.7 g of acid.

Now we know that mass of solution = mass solute + mass solvent

mass solution = 14.7 + 1000 = 1014.7 g of solution

With the density, we can calculate the volume of solution:

d = mass/V - --> V = mass / d

V = 1014.7/1.83 = 553.48 mL of solution

So the 0.15 m, are in 553.48 mL of solution, or 0.55348 L of solution, therefore, the molarity will be:

M = 0.15 / 0.55348 = 0.271 M

Now that we have the molarity, for relation of mole ratio, we know that:

moles A = moles B

we will say "A" is the original solution, and "B" the solution needed.

nA = nB

therefore we know that nA is:

nA = Ma*Va

nB = MB * VB

replacing we have:

0.271Va = 0.080 * 50

Va = 14.76 mL

And this is the volume of acid needed to take to prepare the solution.