The number of moles of water present in a sph | vedclass

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(B) The volume of the spherical droplet is given by $V = \frac{4}{3} \pi r^3$. Given $r = 1.0 \ cm$,so $V = \frac{4}{3} \pi (1.0)^3 = \frac{4 \pi}{3}

Vedclass · Accepted Answer

$\frac{2 \pi}{27}$

Vedclass · Answer

(B) The volume of the spherical droplet is given by $V = \frac{4}{3} \pi r^3$. Given $r = 1.0 \ cm$,so $V = \frac{4}{3} \pi (1.0)^3 = \frac{4 \pi}{3} \ cm^3$. The mass of the water droplet is $m = 	ext{density} 	imes 	ext{volume} = 1.0 \ g \ cm^{-3} 	imes \frac{4 \pi}{3} \ cm^3 = \frac{4 \pi}{3} \ g$. The molar mass of water $(H_2O)$ is $M = 18 \ g \ mol^{-1}$. The number of moles $n$ is calculated as $n = \frac{m}{M} = \frac{4 \pi / 3}{18} = \frac{4 \pi}{54} = \frac{2 \pi}{27} \ mol$.

The number of moles of water present in a spherical water droplet of radius $1.0 \ cm$ is,(Given: Density of water in the droplet $= 1.0 \ g \ cm^{-3}$)

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