Question

a. 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?

Answer 1

Answer:

A. 2.139g of KIO3

B. 26.67mL

Explanation:Please see attachment for explanation

Answer 2

(a) The mass of $\rm KIO_3$ has in 0.2 M solution has been 2.14 g.

(b) The volume of sulfuric acid requires has been 26.66 mL.

The molarity of the solution has been given as the moles of solute present in the liter of solution.

Molarity has been expressed as:

$\rm Molarity=\dfrac{Mass}{Molar\;mass}\;\times\;\dfrac{1000}{Volume\;(L)}$

Computation for the mass of $\rm KIO_3$

The molar mass of $\rm KIO_3$ has been 214 g/mol.

(a) Volume of solution, 50 mL

The molarity of the solution has been, 0.2 M.

Substituting the values for mass:

$\rm 0.2=\dfrac{Mass}{214}\;\times\;\dfrac{1000}{50}\\Mass=\dfrac{0.2\;\times\;214}{20}\\Mass=2.14\;g$

The mass of $\rm KIO_3$ has in 0.2 M solution has been 2.14 g.

(b) The volume of solution required has been given by:

$M_1V_1=M_2V_2$

Where, the molarity of a given solution, $M_1=0.15\;\rm M$

The molarity of the required solution, $M_2=0.080 \rm M$

The volume of the required solution, $V_2=\;50\rm mL$

The volume of the given solution required has been:

$0.15\;\times\;V_1=50\;\times\;0.08\;\\V_1=\dfrac{4}{0.15} \\V_1=26.66\;\rm mL$

The volume of sulfuric acid requires has been 26.66 mL.

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