Question

the heat of vaporization of water at 100*c is 40.66 kj/mol. calculate the quantity of heat that is absorbed/released when 9.00 g of steam condenses to liquid water at 100*c

Answer 1

Concepts and reason

The concept used here to solve this problem is based on the heat of vaporization.

Firstly, the number of moles of steam is calculated using the given mass. After that, the amount of heat released/absorbed is calculated using the heat of vaporization.

Fundamentals

The number of moles is calculated as shown below.

$n = \frac{m}{M}$ ……(1)

Here, $n$ is the number of moles, $m$ is the given mass and $M$ is the molar mass.

The heat of vaporization:

It is the amount of energy needed to change one gram of liquid to a gas at a constant temperature.

$\Delta {H_{{\rm{vap}}}} = \frac{q}{n}$ …… (2)

Here, $q$ is the heat released or absorbed and $n$ is the number of moles.

Substitute the value of $m$ as ${\rm{9}}{\rm{.00 g}}$ and $M$ as ${\rm{18}}{\rm{.02 g/mol}}$ in the equation (1).

$\begin{array}{l}\\n = \frac{{{\rm{9}}{\rm{.00 }}\cancel{{\rm{g}}}}}{{{\rm{18}}{\rm{.02 }}\cancel{{\rm{g}}}{\rm{/mol}}}}\\\\n = {\rm{0}}{\rm{.4994 mol}}\\\end{array}$

Substitute the value of $\Delta {H_{{\rm{vap}}}}$ as ${\rm{40}}{\rm{.66 kJ/mol}}$ and $n$ as ${\rm{0}}{\rm{.4994 mol}}$ in the equation (2).

$\begin{array}{c}\\{\rm{40}}{\rm{.66 kJ/mol}} = \frac{q}{{{\rm{0}}{\rm{.4994 mol}}}}\\\\q = {\rm{40}}{\rm{.66 kJ/}}\cancel{{{\rm{mol}}}} \times {\rm{0}}{\rm{.4994 }}\cancel{{{\rm{mol}}}}\\\\q = 20.3{\rm{ kJ}}\\\end{array}$

Ans:

The amount of heat released of the steam condenses to liquid is ${\bf{20}}{\bf{.3 kJ}}$ .