Question

question 1 of 25 what volume of water can be boiled by 3.0 kj of energy? (refer to table of constants for water.) a. 3.0 kj×\frac{1 mol}{6.03 kj}×18.02 g/mol×\frac{1 ml}{1 g}=9.0 ml b. 3.0 kj×\frac{1 mol}{4.186 kj}×18.02 g/mol×\frac{1 ml}{1 g}=13 ml c. 3.0 kj×\frac{1 mol}{(-285.83 kj)}×18.02 g/mol×\frac{1 ml}{1 g}=0.19 ml d. 3.0 kj×\frac{1 mol}{40.65 kj}×18.02 g/mol×\frac{1 ml}{1 g}=1.3 ml

Explanation:

Step1: Identify heat - of - vaporization of water

The heat of vaporization of water is the amount of energy required to convert 1 mole of water from liquid to gas at its boiling point. The heat of vaporization of water is 40.65 kJ/mol.

Step2: Calculate moles of water

We know the energy input ($Q = 3.0$ kJ) and the heat of vaporization ($\Delta H_{vap}=40.65$ kJ/mol). Using the formula $n=\frac{Q}{\Delta H_{vap}}$, where $n$ is the number of moles. So $n = 3.0\ kJ\times\frac{1\ mol}{40.65\ kJ}$.

Step3: Calculate mass of water

The molar mass of water ($H_2O$) is $M = 18.02$ g/mol. The mass $m=n\times M$. Substituting $n$ from step 2, we get $m=(3.0\ kJ\times\frac{1\ mol}{40.65\ kJ})\times18.02\ g/mol$.

Step4: Calculate volume of water

The density of water is $
ho = 1\ g/mL$. Using the formula $V=\frac{m}{
ho}$, and since $
ho = 1\ g/mL$, $V=(3.0\ kJ\times\frac{1\ mol}{40.65\ kJ})\times18.02\ g/mol\times\frac{1\ mL}{1\ g}\approx1.3\ mL$.

Answer:

D. 3.0 kJ×$\frac{1\ mol}{40.65\ kJ}$×18.02 g/mol×$\frac{1\ mL}{1\ g}$ = 1.3 mL