By Pleistocene ice age is meant the glacial periods that appeared in the Pleistocene. The Pleistocene is the first epoch of the Quaternary. It is an epoch when there was a so called ''ice age'' on the Earth, or rather a glacial period. During this ice age the planet had much lower temperatures on a global scale. The climate was also much drier. Lot of ocean water was frozen in the ice sheets that were stretching deep into the North American and Eurasian continents, which resulted in much lower sea levels that today as well. The places further north than 40 degrees of latitude were almost exclusively covered with ice, so life was almost impossible apart from some coastline places.

7 0

4 years ago

A solution was prepared by dissolving 195.0 g of KCl in 215 g of water. Calculate the mole fraction of KCl. (The formula weight

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Answer:

Approximately $0.180$ .

Explanation:

The mole fraction of a compound in a solution is:

$\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}$ .

In this question, the mole fraction of $\rm KCl$ in this solution would be:

$\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}$ .

This solution consist of only $\rm KCl$ and water (i.e., $\rm H_2O$ .) Hence:

$\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}$ .

From the question:

Mass of $\rm KCl$ : $m(\mathrm{KCl}) = 195.0\; \rm g$ .
Molar mass of $\rm KCl$ : $M(\mathrm{KCl}) = 74.6\; \rm g \cdot mol^{-1}$ .

Mass of $\rm H_2O$ : $m(\mathrm{H_2O}) = 215\; \rm g$ .
Molar mass of $\rm H_2O$ : $M(\mathrm{H_2O}) = 18.0\; \rm g\cdot mol^{-1}$ .

Apply the formula $\displaystyle n = \frac{m}{M}$ to find the number of moles of $\rm KCl$ and $\rm H_2O$ in this solution.

$\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}$ .

$\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}$ .

The molar fraction of $\rm KCl$ in this solution would be:

$\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}$ .

(Rounded to three significant figures.)

8 0

3 years ago