Question #1f4d1

Question

Question #1f4d1

1 Answer

Impact of this question

4785 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-22T00:03:23

About $34 g$ $"34 g"$ .

Explanation:

You can solve this problem by looking at the solubility graph for potassium nitrate, ${KNO}_{3}$ $"KNO"_3$ .

![www.mts.net]()

For starters, the temperature is given to you in Kelvin, so convert it to degrees Celsius

$t [^{\circ} C] = 313 K - 273.15 K \approx 40^{\circ} C$ $t[""^@"C"] = "313 K" - "273.15 K" ~~ 40^@"C"$

Now, according to the solubility graph, potassium nitrate has a solubility of about ${67 g/100 g H}_{2} O$ $"67 g/100 g H"_2"O"$ . This means that in order to have a saturated solution of potassium nitrate at $40^{\circ} C$ $40^@"C"$ , you must dissolve about $67 g$ $"67 g"$ of this salt in $100 g$ $"100 g"$ of water.

In other words, an aqueous solution of potassium nitrate will contain a maximum of $67 g$ $"67 g"$ of dissolved potassium nitrate for every $100 g$ $"100 g"$ of water at $40^{\circ} C$ $40^@"C"$ .

You can now use this solubility to figure out how many grams of potassium nitrate can be dissolved in $50 g$ $"50 g"$ of water at this temperature to have a saturated solution

$50 color(red)(cancel(color(black)("g H"_2"O"))) * "67 g KNO"_3/(100color(red)(cancel(color(black)("g H"_2"O")))) = "34 g KNO"_3$

I'll leave the answer rounded to two sig figs.

Science

Math

Humanities

... and beyond

Question #1f4d1

1 Answer

Explanation:

Related questions

Impact of this question