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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $= 1.0 \ g \ cm^{-3}$)

Similar Questions

How many oxygen atoms are present in $4.4 \ g$ of $CO_2$?

What is the molarity of $H_2O$ in pure water?

In a flask,the weight ratio of $CH_{4(g)}$ and $SO_{2(g)}$ at $298 \ K$ and $1 \ bar$ is $1:2$. The ratio of the number of molecules of $SO_{2(g)}$ and $CH_{4(g)}$ is

How many moles of magnesium phosphate,$Mg_{3}(PO_{4})_{2}$ will contain $0.25 \ mol$ of oxygen atoms?

Which gas among the following contains the maximum number of molecules at $STP$? (Molar masses in $g \ mol^{-1}$: $CO_{2}=44, Ar=39.9, CH_{4}=16, O_{2}=32$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module