A) \[1.5\times {{10}^{25}}\]
B) \[2.00\times {{10}^{23}}\]
C) \[3.4\times {{10}^{25}}\]
D) \[4.0\times {{10}^{24}}\]
Correct Answer: A
Solution :
[a] \[{{q}_{p}}=\Delta H={{C}_{p}}dT\] \[\Rightarrow {{q}_{p}}=75.32\frac{J}{K\,mol}\times (299-298)K\] \[\Rightarrow {{q}_{p}}=75.32\frac{J}{K\,mol}\] For \[180\text{ }kg\] of water, no. of moles of water \[=\frac{180\times {{10}^{3}}g}{18g/mol}={{10}^{4}}g\,moles\,\] \[{{q}_{p}}=75.32\frac{J}{mol}\times {{10}^{4}}moles\] \[=753.2\times {{10}^{3}}J=753.2\,kJ\] \[\Delta H\] for\[ATP=7kcal/mol\] \[=7\times 4.184\,kJ/mol\] \[=29.2kJ/mol\] \[6.022\times {{10}^{23}}\] molecules of ATP produce \[=29.2\text{ }kJ\] \[29.2\text{ }kJ\] produced from \[6.022\times {{10}^{23}}\] molecules \[753.2\text{ }kJ\] produced from \[6.022\times {{10}^{23}}\times \frac{75.8}{29.2}\] \[=1.5\times {{10}^{25}}\] molecules
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