An empty relative density bottle weighs 40g. The bottle weighs 80g when filled with methylated spirit. And 90g when filled with water. Find the density of the methylated spirit. Take density of water to be 1000 `kg*m^(-3).`

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I think that the volumes of spirit and water were the same, and probably equal to the inner volume of a bottle. Denote this volume as `V_b.`

A volume `V,` a mass `m` and a density `rho` are bounded by the relation `rho=m/V.` In particular,

`rho_s=m_s/V_b` and `rho_w=m_w/V_b,`

where...

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Hello!

I think that the volumes of spirit and water were the same, and probably equal to the inner volume of a bottle. Denote this volume as `V_b.`

A volume `V,` a mass `m` and a density `rho` are bounded by the relation `rho=m/V.` In particular,

`rho_s=m_s/V_b` and `rho_w=m_w/V_b,`

where `rho_s` is the density of spirit, `m_s` is its mass, `rho_w` is the density of water and `m_w` is its mass. Denote also the mass of a bottle as `m_b.`

The masses are actually known: `m_s=80g-m_b=40g` and `m_w=90g-m_b=50g.`

So we have two (linear) equations and two unknowns, `rho_s` and `V_b.` From the second equation `V_b=m_w/rho_w,` substitute it into the first equation and obtain:

`rho_s=m_s/V_b=(m_s/m_w)*rho_w.`

We can use any units for `rho_w` here because `m_s/m_w` is dimensionless. Therefore:

`p_s` = (40/50)*1000 = 800 `((kg)/m^3)`.

 

 

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