Respuesta :
Answer:
The correct answer to the following question will be Option A (-28.36 KJ). The further explanation is given below.
Explanation:
The given values are:
Sodium's mass = 8 grams
Water's volume = 227 cm³
Molar mass = 23 g/mol
Density of water = 1 g/cm³
Specific heat capacity ([tex]C_{w}[/tex]) = 4.18 J/Kg
As we know,
⇒ [tex]Moles \ of \ Na =\frac{mass \ of \ Na}{Molar \ mass \ of \ Na}[/tex]
On putting the estimated values, we get
⇒ [tex]=\frac{8}{23}[/tex]
⇒ [tex]=0.348 \ mol[/tex]
Now, Water's mass will be:
⇒ [tex]M_{w} = density\times volume[/tex]
[tex]=1\times 227[/tex]
[tex]=227 \ g[/tex]
Change in temperature will be:
⇒ [tex]\Delta T=(308.4-298)K[/tex]
[tex]=10.4 \ K[/tex]
Heat released will be:
⇒ [tex]q=M_{w}\times C_{w}\times \Delta T[/tex]
On substituting the estimated values, we get
[tex]=227\times 4.18\times 10.4[/tex]
[tex]=9.87\times 10^3[/tex]
[tex]=9.87 \ KJ[/tex]
So that the change in the solution of Na will be:
⇒ [tex]\Delta H=\frac{-q}{Moles \ of \ Na}[/tex]
[tex]=\frac{-9.87}{0.348}[/tex]
[tex]=-28.36 \ KJ/mol[/tex]
