CHM 130LL Experiment 11 PRE-LAB QUESTIONS 1. Identify solute and sol Name: Class
ID: 1033635 • Letter: C
Question
CHM 130LL Experiment 11 PRE-LAB QUESTIONS 1. Identify solute and sol Name: Class number: solute: solvent: 2. For a 12.6% NaCl solution, calculate the solubility of Naci in 100 g of water. Calculation: 3. Which of the following solutions cannot dissolve any more solutes? i) dilute solution ii) unsaturated solution ii) saturated solution iv) none of these 4. Which of the following is a strong acid? i) acetic acid iii) nitric acid ii) sulfurous acid iv) carbonic acid 5. Which of the following is an insoluble salt (-insoluble ionic compound)? See solubility table given in the experiment 9. i) BaSO4 ii) Naso. ii) Ca(NO, iv) CO 6. Circle the electrolyte(s). (Hint: See Table 11.1 and solubility table.)Explanation / Answer
[1] In a salt solution solute is salt and solvent is water. A salt solution is prepared by dissolving some amount of salt into some amount of water.
Solute is Salt
Solvent is Water
[2] 12.6 % NaCl solution means if one takes 100 mL of solution, it contains 12.6 % of NaCl.
12.6 % of 100 is 12.6
So, 12.6 gram of NaCl is present in 100 mL of solution.
Thus, solubility of NaCl is 12.6 g/100mL
[3] the answer is (ii) saturated solution
A saturated solution can not dissolve any more solute because a saturated solution is prepared by dissolving the maximum amount of solute in a given amount of solvent. Since, the maximum amount of solute which can be dissolved is already dissolved in solution, now one can not dissolve any more solute in it.
[4] the answer is (iii) Nitric acid
Nitric acid completely dissociates in water to furnish H+ ions. While other also gives H+ ions but they does not dissociate completely. Hence, nitric acid is strong acid.
[5] the answer is (i) BaSO4
Reason: Check the solubility table. You will find that the solubility product is of the order of 10-10. And, the solubility is very low (~0.00025 g/100ml).
Also, CO2 is not a salt. While Na2SO4 and Ca(NO3)2 have high solubility.
[6] KCl, NaCl and HCl are elctrolytes. C6H12O6 is not an electrolyte.
An electrolyte is one which gives cations and anions
KCl (aq) ----> K+(aq) + Cl-(aq)
NaCl (s) ----> Na+ + Cl- (it exist as ion in solid)
HCl (aq) ----> H+(aq) + Cl-(aq)
C6H12O6 simply dissolves in water, and does produce any ion.