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Question: The relative density of mercury is 13.6, if the volume taken of water is 10 times the volume of merc...

The relative density of mercury is 13.6, if the volume taken of water is 10 times the volume of mercury, then find the ratio of the mass of mercury to the mass of water taken.

Explanation

Solution

Hint: Apply the formula for relative density (ρr)\left( {{\rho }_{r}} \right) i.e., ρr=ρHgρw{{\rho }_{r}}=\dfrac{{{\rho }_{Hg}}}{{{\rho }_{w}}}
Where, ρHg{{\rho }_{Hg}} is the density of mercury and, ρw{{\rho }_{w}} is the density of water at 4C4{}^\circ C

Complete step by step answer:

Given that the relative density of mercury is 13.6
So, ρr=ρHgρw=13.6{{\rho }_{r}}=\dfrac{{{\rho }_{Hg}}}{{{\rho }_{w}}}=13.6
Now it is also given that the volume taken of water is 10 times the volume of mercury.
Therefore, applying ρr=ρHg.ρw=mHgVHgmwVw{{\rho }_{r}}=\dfrac{{{\rho }_{Hg.}}}{{{\rho }_{w}}}=\dfrac{\dfrac{{{m}_{Hg}}}{{{V}_{Hg}}}}{\dfrac{{{m}_{w}}}{{{V}_{w}}}},
13.6=ρHg.ρw=mHgVHgmw10VHg13.6=\dfrac{{{\rho }_{Hg.}}}{{{\rho }_{w}}}=\dfrac{\dfrac{{{m}_{Hg}}}{{{V}_{Hg}}}}{\dfrac{{{m}_{w}}}{10{{V}_{Hg}}}}
On solving,
mHgmw=13.610\dfrac{{{m}_{Hg}}}{{{m}_{w}}}=\dfrac{13.6}{10}
mHgmw=1.36\dfrac{{{m}_{Hg}}}{{{m}_{w}}}=1.36

Therefore, the ratio of the mass of mercury to the mass of water taken will be, 1.36

Additional Information:

Density: The density of a substance is its mass per unit volume.
Density( !!ρ!! )=mass(m)volume(V)\text{Density( }\\!\\!\rho\\!\\!\text{ )=}\dfrac{\text{mass(m)}}{\text{volume(V)}}
S.I. unit of density is kg/m3\text{kg/}{{\text{m}}^{\text{3}}}

Relative Density: Relative density (R.D.) of a substance is the ratio of the density of the substance to the density of water at 4°C. i.e., ρr=ρHgρw{{\rho }_{r}}=\dfrac{{{\rho }_{Hg}}}{{{\rho }_{w}}}
It is also defined as the ratio of the mass of the substance to the mass of an equal volume of water at 4°C. R.D. has no unit.
R.D. of a solid substance is given by Archimedes’ principle as,
R.D.( !!ρ!! r)=Weight of the solid in air(Weight of the body in the air - Weight of the body in the water)\text{R}\text{.D}\text{.(}{{\text{ }\\!\\!\rho\\!\\!\text{ }}_{\text{r}}}\text{)=}\dfrac{\text{Weight of the solid in air}}{\left( \text{Weight of the body in the air - Weight of the body in the water} \right)}

Note: Students should understand the whole concept of density and relative density both, so that they can do this type of question easily. Along with this, students should memorise the formula for density and relative density. There is one more thing to keep in mind that relative density is the ratio of the same quantities so relative density has no unit.