Question

A droplet of mercury has a radius of 5mm. how many mercury atoms are there in the
droplet?( assume the droplet is spherical And use M(molar mass) = 200gram and density
of Mercury = 13600kg/m 3 )
Droplet volume =4/3pir 3 =5.2x10 -7 m 3
Droplet mass (m) = density x volume =7.1x 10 -3 kg
Mass of mercury atom (m p ) =M/N A = 3.3x 10 -25 kg
Number of atoms in droplet (N) = m/m p

Answer 1

Given:

Radius of drop= 5mm= 5 X 10⁻³m

Density  of Mercury = 13600kg m⁻³

M(molar mass) of Mercury = 200g= 0.2kg

To find:

No. of mercury atoms in the drop

Solution:

Volume of sphere= $\frac{4}{3} \pi r^{3}$

Volume of drop= 5.2x10⁻⁷m³

Mass of Hg in the drop= Density X Volume

= 13600kg m⁻³ X 5.2x10⁻⁷m³

=707.2 X 10⁻⁵kg

No. of moles of Hg in given mass=  $\frac{707.2 X 10^{-5} }{0.2}$

No. of Hg atoms = $\frac{707.2 X 10^{-5} }{0.2}$ X Nₐ    (Nₐ=6.022X10²³)

= 21293 X 10¹⁸

Hence, there are 2.12 X 10²² atoms in the given drop.

Answer 2

Answer:

A droplet of mercury has a radius of 5mm.  ( assume the droplet is spherical And use M(molar mass) = 200gram and density  of Mercury 13600kg/m3

The number of atoms in the mercury droplet is $2.3*10^{22}$

Explanation:

The mass of mercury droplet =$density *volume$

From the question, the density of mercury is given as, ρ=$13600kg/m^{3}$

assume the droplet is spherical to find the volume therefore volume of the droplet

$V=\frac{4}{3} \pi r^{3}$ (volume of a sphere)

where $r=5mm=0.005m$ the radius of the droplet

substituting the values we get volume $V=\frac{4}{3}\pi (0.005)^{3} =5.23*10^{-7} m^{3}$

mass of mercury $m=13600*5.23*10^{-7}=711.28*10^{-5}kg$

To find the number of atoms in a given sample first we need to find the number of moles in the sample.

The number of moles in mercury droplet $n=mass of given sample/molar mass$

the molar mass of mercury =$200g=0.2kg$

$n=711.28*10^{-5}/0.2=3556.4$ moles

we know $1 mole =6.626*10^{23} atoms$ (Avogadro's number)

Here atoms in the mercury droplet are =$711.28*10^{-5} *6.26*10^{23} =2.3*10^{22}$ atoms.