Question

A spherical ball of radius 7 cm contains 56% iron, if density is 1.4 g/cm cube, number of moles fe present approximately is

Answer 1

Find the volume of the cube

Volume of a sphere = 4/3πr³

      volume of this sphere = 4/3 × 22/7 × 7 

                                            = 4/3 × 22/7 × 7

                                            = 29.33 cm³

Find mass of the cube using density

density is 1.4g/cm³

That means mass of the sphere = 1.4 × 29.33
                                                    =   41.067 g


Iron makes up 56% of this spherical ball

Mass of the iron in the spherical ball = 56/100 × 41.067 = 22.9973 g
                                                          = 23 g

Calculate moles of Fe present:

Moles = mass/molar mass

mass = 23g
molar mass = 56g/mol

moles= 23/56

          = 0.4107 moles.

The moles of Fe in the spherical ball is 0.4107 moles

Answer 2

Answer:

Step-by-step explanation:1st of all let us calculate volume of sphere

$(4/3) \pi r^{3}$$=(4/3)\pi 7^{3}$

volume= 1437.33 cc

only 56% of whole volume is of iron then

volume of iron=804.906 cc

NOW,, from

density*volume=mass

1.4*804.906=1126.86 g

THEN,, moles= mass/molar mass

moles=1126.86/56=20.1225≈20 moles