To solve this problem, we need to analyze the information given about the mixture of MCl (a volatile compound) and NaCl, and how they react with AgNO3 to form a precipitate. Let's break down the data step by step to determine the correct statements regarding the ionic mass of M+ and the composition of the mixture.
Understanding the Reaction
When MCl and NaCl are mixed with AgNO3, they react to form silver chloride (AgCl), which is the white precipitate observed. The reactions can be represented as follows:
- MCl + AgNO3 → AgCl + MNO3
- NaCl + AgNO3 → AgCl + NaNO3
Analyzing the First Reaction
From the first part of the problem, we know that 11.2 g of the mixture produced 28.7 g of white precipitate. Since both MCl and NaCl produce AgCl, we can set up the following relationship:
Let x be the mass of MCl and y be the mass of NaCl in the mixture. Thus, we have:
- x + y = 11.2 g (total mass of the mixture)
- AgCl produced = 28.7 g
Calculating Molar Masses
The molar mass of AgCl is approximately 143.32 g/mol. To find the moles of AgCl formed, we can use the formula:
Number of moles of AgCl = mass of AgCl / molar mass of AgCl = 28.7 g / 143.32 g/mol ≈ 0.200 moles
Since each mole of MCl and NaCl produces one mole of AgCl, we can say that:
- moles of MCl + moles of NaCl = 0.200 moles
Analyzing the Second Reaction
Next, we consider the second part of the problem, where heating the same mixture produced a gas that also formed a precipitate when passed into AgNO3, resulting in 14.35 g of AgCl. Using the same calculation:
Number of moles of AgCl = 14.35 g / 143.32 g/mol ≈ 0.100 moles
Setting Up the Equations
Now we have two equations:
- moles of MCl + moles of NaCl = 0.200
- moles of MCl = 0.100 (since only MCl produces gas)
From the second equation, we can substitute into the first:
0.100 + moles of NaCl = 0.200
moles of NaCl = 0.100
Finding the Masses
Now we can find the masses of MCl and NaCl:
- Let the molar mass of MCl = M + 35.5 (where M is the molar mass of M+)
- Let the molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Using the moles calculated:
- Mass of MCl = moles × molar mass = 0.100 × (M + 35.5)
- Mass of NaCl = 0.100 × 58.5 = 5.85 g
Solving for M
Substituting back into the total mass equation:
0.100 × (M + 35.5) + 5.85 = 11.2
0.100M + 3.55 + 5.85 = 11.2
0.100M + 9.4 = 11.2
0.100M = 1.8
M = 18 g/mol
Conclusion on the Options
Now, let's evaluate the options provided:
- (a) Ionic mass of M+ is 18 - **True**
- (b) Mixture has equal mol fraction of MCl and NaCl - **True** (both are 0.100 moles)
- (c) MCl and NaCl are in 1:2 molar ratio - **False** (they are in a 1:1 ratio)
- (d) Ionic mass of M+ is 10 - **False**
Thus, the correct answers are (a) and (b). This analysis shows how stoichiometry and mass relationships can help us deduce the properties of chemical mixtures effectively.
