Answered

### Questions for Freezing Point Depression

#### Table 6.1
\begin{tabular}{|c|c|c|c|c|c|}
\hline Solution & \begin{tabular}{c}
Lowest \\
Temperature [tex]${ }^{\circ} C$[/tex]
\end{tabular} & \begin{tabular}{c}
[tex]$1^{\text {st }}$[/tex] Trial [tex]${ }^{\circ} C$[/tex] \\
Freezing Point
\end{tabular} & \begin{tabular}{c}
[tex]$2^{\text {nd }}$[/tex] Trial [tex]${ }^{\circ} C$[/tex] \\
Freezing Point
\end{tabular} & \begin{tabular}{c}
Average [tex]${ }^{\circ} C$[/tex] \\
Freezing Point
\end{tabular} & [tex]$\Delta T { }^{\circ} C$[/tex] \\
\hline Sucrose & -7.9 & -1.6 & -1.6 & & \\
\hline [tex]$H_2O$[/tex] & -8.0 & +0.0 & +0.0 & & \\
\hline
\end{tabular}

1. Was there supercooling? Would you expect the water or the sugar solution to have the most supercooling (see the Introduction)? Why?

2. The mass of one mole of sugar is [tex]$342 g$[/tex]. You used [tex]$19 g$[/tex] of sugar. Calculate the number of moles of sugar you used.



Answer :

GinnyAnswer
2. The mass of one mole of sugar is [tex]\(342 \text{ g}\)[/tex]. You used [tex]\(19 \text{ g}\)[/tex] of sugar. Calculate the number of moles of sugar you used.

To determine the number of moles of sugar used, we need to use the formula for calculating moles:

[tex]\[ \text{Number of moles} = \frac{\text{Mass of the substance used}}{\text{Molar mass of the substance}} \][/tex]

Given the data:
- The molar mass of sugar (sucrose) is [tex]\(342 \text{ g/mol}\)[/tex].
- The mass of sugar used is [tex]\(19 \text{ g}\)[/tex].

We can substitute these values into the formula:

[tex]\[ \text{Number of moles} = \frac{19 \text{ g}}{342 \text{ g/mol}} \][/tex]

Perform the division:

[tex]\[ \text{Number of moles} \approx 0.05555555555555555 \][/tex]

Therefore, the number of moles of sugar used is approximately [tex]\(0.056 \text{ moles}\)[/tex] (rounded to three significant figures).

Other Questions