Chapter 15, Problem 106A

Interpretation Introduction

Interpretation: To determine the mass and volume of water at STP.

Concept Introduction: Ideal gas law gives a relationship between pressure, temperature, and volume. It can be expressed as $\text{PV=nRT}$ .

Answer to Problem 106A

The mass of water will be ${\text{36×10}}^{\text{-3}}\text{g}$ and volume of the water will be $\text{0}\text{.044 L}$

Explanation of Solution

The given data:

A mole of hydrogen peroxide = ${\text{2×10}}^{\text{-3 }}\text{mole}$

At STP,

Temperature (T) = 273 K

Pressure (P) = 1 atm

According to the equation,

2 moles of hydrogen peroxide decompose to form 2 moles of water.

So, ${\text{2×10}}^{\text{-3 }}\text{mole}$ of hydrogen peroxide decomposes to form ${\text{2×10}}^{\text{-3 }}\text{mole}$ of water.

The mass of the water can be calculated by using the formula:

  $\begin{array}{c}\text{Number}\text{of}\text{moles=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{Mass=}\text{molar}\text{mass×number}\text{of}\text{moles}\\ {\text{Mass=18×2×10}}^{\text{-3}}\text{g}\\ =36\times {\text{10}}^{\text{-3}}\text{g}\end{array}$

Therefore, the mass of the water will be ${\text{36×10}}^{\text{-3}}\text{g}$ .

Volume can be determined by using the ideal gas equation.

According to the ideal gas equation,

  $\text{PV=nRT}$ …(i)

Put the values of the given data in the above equation.

  $\begin{array}{c}\text{PV}=\text{nRT}\\ \text{V=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{{\text{2×10}}^{\text{-3}}\text{×0}\text{.0821×273}}{\text{1}}\\ =\text{0}\text{.044 L}\end{array}$

Therefore, the volume of the water is $\text{0}\text{.044 L}$ .