You have a 0.355 m potassium chloride solution. The solvent is water. What is th
ID: 478539 • Letter: Y
Question
You have a 0.355 m potassium chloride solution. The solvent is water. What is the mass percent of potassium chloride in it? Given the reaction; 2 Al (s) + 6 HNO_3 (aq) rightarrow 2 Al(NO_3)_3 (aq) + 3 H_2 (g) how many mL of 4.72% HNO_3 would be required to react with 0.683 g of Al? The density of the HNO_3 solution is 1.08 g/mL: its solvent is water. A solution is 5.60% methanol by mass. The density of the solution is 0.853 g/mL; its solvent is water. How many grams of methanol and of water would be required to prepare 40.0 mL of solution?Explanation / Answer
1.
As solvent is water Molarity and molality will be same.
We have m = 0.355 m Potassium chloride solution.
To prepare 1 m solution of KCl we require 74.5513 g KCl in 1000 g water.
Grams of KCl = 74.5513 * 0.355 = 26.46 g
Mass percent of KCl = mass of solute/mass of solution * 100 = 26.46/(1000+26.46) *100
= 2.577 %
2.
Wt of Al = 0.683 g
Moles of Al = 0.0253 moles
Moles of HNO3 required are 3 * moles of Al = 0.0253 * 3 = 0.0759 moles
Gram of HNO3 from 0.0759 moles = 4.782 g of HNO3
4.72% HNO3 is 4.72 g HNO3 in 100 g water. But we require 4.782 g of HNO3
Weight of water required is = 4.782*100/4.72 = 101.31 g of water
Total weight = 101.31 + 4.782 = 106.092 g
Volume of HNO3 from density = 106.092/1.08 = 98.23 mL of 4.72% HNO3.
3.
A solution is 5.60% methanol by mass.
We have 5.6 g methanol in 100 g water.
Density of solution is 0.853 g/mL. Therefore, we have 0.853 * 40 mL = 34.12 g of solution.
Now we have 5.60% methanol by mass. Therefore, 34.12 * 0.056 = 1.910 g of methanol
Grams of water = 34.12 – 1.910 = 32.21 g of water
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.