Question

Bond dissociation energy of AB molecule is $300kj/mole$. The number of moles of photons of wavelenght $6625A^o$ required to dissociate $3$ moles of AB molecules is ?

• A. $1$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

Given Bond dissociation energy of $AB$ molecule

$=300$ kJ/mol

$=300\times {10}^3\times 6.023\times {10}^{23}$ J

$=0.498\times {10}^{-18}J⟶\left(1\right)$

Now to find out the number of moles of protons for $3$ moles of $AB$ molecule.

$3\times E=\frac{nhc}{\lambda }$

$n=\frac{3\times E\times \lambda }{hc}$

$=\frac{3\times 0.498\times {10}^{-18}\times 6625\times {10}^{-10}}{6.625\times {10}^{34}\times 3\times {10}^8}$

$=4.98\simeq 5$ mol