Chapter 15, Problem 84A

Agriculture Water is sprayed on oranges during afrosty night. If an average of 11.8 g of water freezes oneach orange, how much heat is released?

Interpretation Introduction

Interpretation:

Heat released is to be calculated.

Concept introduction:

Molar mass: It is the sum of the total mass in grams of all the atoms that make a mole of a particular molecule.

Answer to Problem 84A

Heat released during the process is $\text{-3}\text{.94kJ}$

Explanation of Solution

Molar mass of ${\text{H}}_{\text{2}}^{}\text{O}$.

Since, ${\text{H}}_{\text{2}}^{}\text{O}$ contains two atoms of hydrogen and one atom of oxygen. The molar mass would be

$\text{H=2×1}\text{.0079g/mol}$

$\text{=2}\text{.0158g/mol}$

$\text{O=15}\text{.9994g/mol}$

$\text{Molar}\text{mass=Sum}\text{of}\text{atomic}\text{weight}\text{of}\text{each}\text{element}\text{present}\text{the}\text{compound}$

$\text{=2}\text{.0158g/mol+15}\text{.9994g/mol}$

$\text{=18}\text{.0152g/mol}$

Given information, 11.8g of ${\text{H}}_{\text{2}}^{}\text{O}$

$\text{1mol=18}\text{.0152g/mol}$

$\begin{array}{l}\text{11}\text{.8g}\text{of}{\text{H}}_{\text{2}}^{}\text{O=}\frac{\text{1}}{\text{18}\text{.0152}}\text{.11}\text{.8}\\ \text{=}\frac{\text{11}\text{.8}}{\text{18}\text{.0152}}\\ \text{=0}\text{.65mol}\end{array}$

Take the molar heat of solidification of water as a conversion factor to convert moles of ${\text{H}}_{\text{2}}^{}\text{O}$ to kJ.

Since,

Then, $\begin{array}{l}\text{1mol=6}\text{.01kJ}\\ \text{0}\text{.65}\text{mol=}\frac{\text{6}\text{.01}}{\text{1}}\text{.0}\text{.65}\\ \text{=}\text{-3}\text{.94kJ}\end{array}$

Conclusion

Heat release can be calculated by the molar mass and Hess’s Law.