A spherical ball of radius $7$ cm contains $56 %$ iron. If the density is $1.4 g / c m^{3}$ , number of moles of $F e$ present approximately is :

10

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15

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20

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25

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Solution

The correct option is A $20$
Volume of spherical ball $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times (7)^{3}$
Mass $= V \times d = \frac{4}{3} \times \frac{22}{7} \times (7)^{3} \times 1.4$ g
As the ball contains $56 % F e$ , Mass $= \frac{4}{3} \times \frac{22}{7} \times (7)^{3} \times 1.4 \times \frac{56}{100} g = 1126.4$ g
Number of moles $= \frac{1126.4}{56} = 20$ as molar mass of $F e$ is $56$ g/mol.

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A spherical ball of radius 7 cm contains 56% iron. If the density is 1.4g/cm3, number of moles of Fe present approximately is :

The correct option is A 20Volume of spherical ball=43πr3=43×227×(7)3Mass=V×d=43×227×(7)3×1.4 gAs the ball contains 56%Fe, Mass =43×227×(7)3×1.4×56100g=1126.4 gNumber of moles =1126.456=20 as molar mass of Fe is 56 g/mol.

A spherical ball of radius $7$ cm contains $56 %$ iron. If the density is $1.4 g / c m^{3}$ , number of moles of $F e$ present approximately is :