Question:
How many grams of K2HPO4 (molar mass: 174.18 g/mol) must be dissolved in one liter
of water to:
1. prepare a 0.2 M solution?
2. prepare one liter of a solution containing 100 ppm K?
3. How many grams of P are contained in the solutions from tasks 1 and
2?
1. 0.2 M solution
Molar mass of K2HPO4= 174.18 g/mol
For 1 L of a 1 M solution, 174.18 g would be required.
For 1 L of a 0.2 M solution:
0.2×174.18 g=34.836 g K2HPO4
0.2×174.18 g=34.836 g K2HPO4
Answer 1:
34.84 g
34.84 g
(rounded to two decimal places)
2. 100 ppm K solution
100 ppm K means 100 mg K per liter of solution.
Molar mass of K = 39.10 g/mol
One molecule of K2HPO4 contains two K atoms:
Mass of K in K2HPO4 = 2×39.10=78.20 g K per mole K2HPO4 2×39.10=78.20 g K per mole
K2HPO4.
Mass fraction of K in K2HPO4:
fK=78.20174.18≈0.4488
fK=174.1878.20≈0.4488
To obtain 100 mg K in 1 L, the following mass of K2HPO4 is required:
100 mg K0.4488≈222.82 mg K2HPO4
0.4488100 mg K≈222.82 mg K2HPO4
Answer 2:
0.2228 g
0.2228 g
(or 222.8 mg K2HPO4 per liter for 100 ppm K)
3. Mass of P in each solution
Molar mass of P = 30.97 g/mol
One molecule of K2HPO4contains 1 P atom, therefore:
Mass fraction of P in K₂HPO₄:
fP=30.97174.18≈0.1777
fP=174.1830.97≈0.1777
For solution 1 (0.2 M):
We have 34.836 g K2HPO4 per liter.
Mass of P:
34.836×0.1777≈6.192 g P
34.836×0.1777≈6.192 g P
For solution 2 (100 ppm K):
We have 0.2228 g K₂HPO₄ per liter.
Mass of P:
0.2228×0.1777≈0.03958 g P
0.2228×0.1777≈0.03958 g P
Answer 3:
Solution 1: 6.19 g P6.19 g P
Solution 2: 0.0396 g P0.0396 g P
