A volume of 60.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 17.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l )

The molarity of KOH = 0.86M

Explanation:

From the equation of reaction

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

nA= 1, nB=2

From the question

CA= 1.5M, VA= 17.2ml, VB= 60ml, CB= ?

Applying (CAVA)/(CBVB) = nA/nB

(1.5×17.2)/(CB×60) = 1/2

Simplify

CB= (1.5×17.2×2)/(60×1)

CB = 0.86M

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3