Question

By how much (in picograms) does the mass of 1 mol of ice at $\mathbf{0}\mathbf{°}\mathbf{C}$differ from that of 1 mol of water at 0°C?

Short answer

The difference in mass due to conversion of ice into water is $0.068\text{pg}$.

Step-by-step solution

Identification of given data

The mass of ice and water is $m=1\text{mol}$

Latent heat and internal energy

The latent heat of any substance is the amount of heat required to change the state of unit amount of substance without any variation in temperature of substance. The difference in mass due to change in ice into water is found by dividing latent heat of ice by internal energy of ice.

Determination of difference in mass

The difference in mass due to conversion from ice into water is given as:

$\Delta m=\frac{H}{c^2}$

Here is the mass of water molecule and its $18\text{g}/\text{mol}$, C is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$, H is the latent heat of ice and its value is $6000\text{J}/\text{mol}$.

Substitute all the values in the equation.

$$\begin{gathered} \Delta m=\frac{\left(18\text{g}\right)\left(334\text{J}/\text{g}\right)}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2} \\ \Delta m=0.068\times {10}^{-12}\text{g} \\ \Delta m=\left(0.068\times {10}^{-12}\text{g}\right)\left(\frac{1\text{pg}}{{10}^{-12}\text{g}}\right) \\ \Delta m=0.068\text{pg} \end{gathered}$$

Therefore, the difference in mass due to conversion of ice into water is $0.068\text{pg}$.