Answer:
[tex]m_w=5.3248\times 10^{23}\ kg[/tex]
Explanation:
According to the given reaction,
heat released by thecombustion of 2 molecules of Methanol, [tex]\Delta H=1277000\ J[/tex]
we know that molecular mass of Methanol, [tex]m=32\ g.mol^{-1}[/tex]
∴12 gram of methanol = [tex]\frac{3}{8}\ moles[/tex]
we know 1 mole = [tex]6.023\times 10^{23}\ molecules[/tex]
so,
[tex]\rm \frac{3}{8} \ moles=2.258\times 10^{23}\ molecules[/tex]
Heat from the combustion of [tex]2.258\times 10^{23}\ molecules[/tex]:
[tex]Q=1277000\times \frac{2.258\times 10^{23}}{2}[/tex]
[tex]Q=1442132.0625\times 10^{23}\ J[/tex]
Now the mass of water that can be heated from 23.5°C to 88.2°C :
[tex]Q=m_w.c_w.\Delta T[/tex]
[tex]1442132.0625\times 10^{23}=m_w.4186\times (88.2-23.5)[/tex]
[tex]m_w=5.3248\times 10^{23}\ kg[/tex]
