Question #06568

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Question #06568

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Answer 1 · 2017-04-10T10:43:24

$the mass of bottle m_{bottle} = 36.19 g$ $"the mass of bottle " m_("bottle") =36.19 g$

Explanation:

$m_{bottle} : mass of bottle$ $m_("bottle"):"mass of bottle"$
$m_{water} : mass of water$ $m_("water"):"mass of water"$

$m_{bottle} + m_{water} = 60 g$ $m_("bottle")+m_("water")=60g$

$m_{water} = 60 - m_{bottle}$ $color(green)(m_("water")=60-m_("bottle"))$

$m_{mercury} : mass of mercury$ $m_("mercury"):"mass of mercury"$

$m_{mercury} + m_{bottle} = 360$ $m_("mercury")+m_("bottle")=360$

$m_{mercury} = 360 - m_{bottle}$ $color(red)(m_("mercury")=360-m_("bottle"))$

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

$V_{water} : Volume of water$ $V_("water"):"Volume of water"$

The volume of water and bottle are equal.

$V_{water} = V_{bottle}$ $V_("water")=V_("bottle")$

$V_{mercury} : Volume of mercury$ $V_("mercury"):"Volume of mercury"$

$V_{mercury} = V_{bottle} = V_{water}$ $V_("mercury")=V_("bottle")=V_("water")$

$V_{mercury} = 60 - m_{bottle}$ $V_("mercury")=color(green)(60-m_("bottle"))$

$d_{mercury} : density of mercury$ $d_("mercury"):"density of mercury"$

$d_{mercury} = \frac{m_{mercury}}{V_{mercury}}$ $d_("mercury")=(m_("mercury"))/(V_("mercury"))$

$d_{mercury} = 13.6 \frac{g}{c m^{3}}$ $d_("mercury")=13.6" " g/(cm^(3))$

$13.6 = \frac{360 - m_{bottle}}{60 - m_{bottle}}$ $13.6=(color(red)(360-m_("bottle")))/(color(green)(60-m_("bottle")))$

$13.6 (60 - m_{bottle}) = 360 - m_{bottle}$ $13.6(color(green)(60-m_("bottle")))=color(red)(360-m_("bottle"))$

$816 - 13.6 m_{bottle} = 360 - m_{bottle}$ $816-13.6m_("bottle")=360-m_("bottle")$

$816 - 360 = 13.6 m_{bottle} - m_{bottle}$ $816-360=13.6m_("bottle")-m_("bottle")$

$456 = 12.6 m_{bottle}$ $456=12.6m_("bottle")$

$m_{bottle} = \frac{456}{12.6}$ $m_("bottle")=456/(12.6)$

$m_{bottle} = 36.19 g$ $m_("bottle")=36.19g$

Answer 2 · 2017-04-10T10:56:14

Let inner volume of the bottle be $v$ $v$ $c m^{3}$ $cm^3$ and mass of empty bottle be $m$ $m$ g

So by the problem

$m + mass of v c m^{3} water = 60 g$ $m +"mass of "vcm^3" water"=60g$

$mass of v c m^{3} water = (60 - m) g$ $"mass of "vcm^3" water"=(60 -m)g$

Again

$m + mass of v c m^{3} of Hg = 360 g$ $m+"mass of "vcm^3" of Hg"=360g$

$mass of v c m^{3} of Hg = (360 - m) g$ $"mass of "vcm^3" of Hg"=(360-m)g$

Now we know

$\frac{mass of v c m^{3} of Hg}{mass of v c m^{3} of water} = sp.gravity of Hg$ $("mass of "vcm^3" of Hg")/ ("mass of "vcm^3" of water")="sp.gravity of Hg"$

$\Rightarrow \frac{mass of v c m^{3} of Hg}{mass of v c m^{3} of water} = 13.6$ $=>("mass of "vcm^3" of Hg")/ ("mass of "vcm^3" of water")=13.6$

$\Rightarrow \frac{360 - m}{60 - m} = 13.6$ $=>(360-m)/ (60-m)=13.6$

$\Rightarrow (360 - m) = (60 - m) \times 13.6$ $=>(360-m)=(60-m)xx13.6$

$\Rightarrow 360 - m = 816 - 13.6 m$ $=>360-m=816-13.6m$

$\Rightarrow 12.6 m = 816 - 360 = 456$ $=>12.6m=816-360=456$

$\Rightarrow m = \frac{456}{12.6} \approx 36.2 g$ $=>m=456/12.6~~36.2g$

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Question #06568

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