Question

A piece of lead weighs 100 g in water and if the density of lead is 11g/cm³ then find the weight of lead in air.

Answer 1

Hey mate
your answer is here
first find volume if lead

V = mass/density

V = 100/11 cm^3

we know that density of air is 1.225 kg/m^3

so,
mass of lead in the air

M = (100/11)/(1000000)×1.225

M = 0.0111 gm

Hope this help you

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Answer 2

Given :

Weight of lead in weight , m = 100 g .

Density of lead , $d = 11\ g/cm^3$ .

To Find :

The weight of lead in air.

Solution :

Density of air is ,

$\rho=1.225\ kg/m^3=\dfrac{1.225\times 1000}{100^3}\ g/cm^3\\\\\rho=1.225\times 10^{-3}\ g/cm^3$

Volume of lead :

$V=\dfrac{m}{d}\\\\V=\dfrac{100}{11}\ cm^3\\\\V=9.09\ cm^3$

Now , weight in air is given by :

$W={\rho}\times{V}\\\\W=1.225\times 10^{-3}\times 9.09\ g\\\\W=0.011\ g$

Therefore , weight of lead in air is 0.011 g .

Hence , this is the required solution .