Question

Question #06568

Answer 1

`"the mass of bottle " m_("bottle") =36.19 g`

Explanation:

`m_("bottle"):"mass of bottle"`
`m_("water"):"mass of water"`

`m_("bottle")+m_("water")=60g`

`color(green)(m_("water")=60-m_("bottle"))`

`m_("mercury"):"mass of mercury"`

`m_("mercury")+m_("bottle")=360`

`color(red)(m_("mercury")=360-m_("bottle"))`

**Since the density of water is 1,the mass and volume of water are equal. **

`V_("water"):"Volume of water"`

The volume of water and bottle are equal.

`V_("water")=V_("bottle")`

`V_("mercury"):"Volume of mercury"`

`V_("mercury")=V_("bottle")=V_("water")`

`V_("mercury")=color(green)(60-m_("bottle"))`

`d_("mercury"):"density of mercury"`

`d_("mercury")=(m_("mercury"))/(V_("mercury"))`

`d_("mercury")=13.6" " g/(cm^(3))`

`13.6=(color(red)(360-m_("bottle")))/(color(green)(60-m_("bottle")))`

`13.6(color(green)(60-m_("bottle")))=color(red)(360-m_("bottle"))`

`816-13.6m_("bottle")=360-m_("bottle")`

`816-360=13.6m_("bottle")-m_("bottle")`

`456=12.6m_("bottle")`

`m_("bottle")=456/(12.6)`

`m_("bottle")=36.19g`

Answer 2

Let inner volume of the bottle be `v` `cm^3` and mass of empty bottle be `m` g

So by the problem

`m +"mass of "vcm^3" water"=60g`

`"mass of "vcm^3" water"=(60 -m)g`

Again

`m+"mass of "vcm^3" of Hg"=360g`

`"mass of "vcm^3" of Hg"=(360-m)g`

Now we know

`("mass of "vcm^3" of Hg")/ ("mass of "vcm^3" of water")="sp.gravity of Hg"`

`=>("mass of "vcm^3" of Hg")/ ("mass of "vcm^3" of water")=13.6`

`=>(360-m)/ (60-m)=13.6`

`=>(360-m)=(60-m)xx13.6`

`=>360-m=816-13.6m`

`=>12.6m=816-360=456`

`=>m=456/12.6~~36.2g`