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Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Answer:
The correct answer is - yes, 4.57 g of solute per 100 ml of solution
Explanation:
The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.
Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.
Then solublity will be :
32. g solute / 700 ml solution = y / 100 ml solution
⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.
Thus, the answer is 4.57 g of solute per 100 ml of solution.
