A spherical ball of radius 7 cm contains 56 | vedclass

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(B) The volume of the spherical ball is given by $V = \frac{4}{3} \pi r^3$. Substituting $r = 7 \ cm$, $V = \frac{4}{3} 	imes \frac{22}{7} 	imes 7^3

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The volume of the spherical ball is given by $V = \frac{4}{3} \pi r^3$. Substituting $r = 7 \ cm$, $V = \frac{4}{3} 	imes \frac{22}{7} 	imes 7^3 = \frac{4}{3} 	imes 22 	imes 49 \approx 1437.33 \ cm^3$. The mass of the ball is $m = V 	imes d = 1437.33 \ cm^3 	imes 1.4 \ g \ cm^{-3} \approx 2012.26 \ g$. The mass of iron in the ball is $56\%$ of the total mass: $m_{	ext{Fe}} = 2012.26 	imes 0.56 \approx 1126.87 \ g$. The number of moles of iron is $n = \frac{m_{	ext{Fe}}}{	ext{atomic mass}} = \frac{1126.87}{56} \approx 20.12 \ mol$. Thus, the number of moles of iron is approximately $20.1 \ mol$. Therefore, option $(B)$ is correct.

$A$ spherical ball of radius $7 \ cm$ contains $56 \ w\%$ iron. Find out the number of moles of iron present approximately in the ball (in $.1$)? $(d = 1.4 \ g \ cm^{-3}; \text{atomic mass} = 56 \ g \ mol^{-1})$

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