Calculating the Concentration of N/5 HCl After Removal of 3.65g

In summary: Let's just go back to Molality.At this point you should know about Molality. There should be an equation for it (Molality) in your textbook... I think that Borek wants you to look it up rather than me tell you what the equation is. You'll really need to memorize what the equation is anyway (don't worry, it's straight forward).
  • #1
xiphoid
57
0

Homework Statement


On boiling 1litre of N/5 HCl, the volume of the aqueous solution decreases to 250ml. If 3.65g of HCl is removed from solution, then the concentration of resulting solution becomes?
N/5
N/2.5
N/20
N/10

Homework Equations


HCl=36.5g/ mol


The Attempt at a Solution


no idea
 
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  • #2
Use concentration definition. You start with 1L of solution - how much solvent? How many moles of HCl? How much solvent after the boiling? How much HCl?
 
  • #3
At this point you should know about Molality.
There should be an equation for it (Molality) in your textbook... I think that Borek wants you to look it up rather than me tell you what the equation is. You'll really need to memorize what the equation is anyway (don't worry, it's straight forward).
 
  • #4
ohms law said:
At this point you should know about Molality.
There should be an equation for it (Molality) in your textbook... I think that Borek wants you to look it up rather than me tell you what the equation is. You'll really need to memorize what the equation is anyway (don't worry, it's straight forward).

not molality (m) but molarity (M). There is a subtle, but important difference. And I do not think that that is where your problem lies because normality (N) is a "drop in" substitute for molarity (M) for monofunctional reagents.

"Equivalent weights" and "Normality" are no longer used by chemists (since a IUPAC ruling in the mid 1960s), but they have lingered on in engineering and in sections of the chemical industry. They become problematic because of ambiguities, where, for example, nitric acid as oxidant may have functionalities of 1, 3, 4, or 5, for its various oxidation reactions, meaning that the exact same solution could be regarded as 0.1 N, 0.3 N, 0.4 N, or 0.5 N!
 
  • #5
oops, sorry. That's what I meant (Molarity). Damn.
I've never heard of "normality" before.

Molality (m=mol/kg) is what we were doing about two weeks ago, so... *shrug*
 

Related to Calculating the Concentration of N/5 HCl After Removal of 3.65g

1. How do you calculate the concentration of N/5 HCl after removal of 3.65g?

To calculate the concentration of N/5 HCl, you will need to know the number of moles of HCl present. To determine this, you can use the formula: moles = mass/molar mass. In this case, the molar mass of HCl is 36.46 g/mol. So, if 3.65g of HCl was removed, the number of moles would be 3.65g/36.46 g/mol = 0.1 mol. Since N/5 HCl means 1/5 of a normal solution, the concentration would be 0.1 mol/5 = 0.02 M.

2. What is the molar mass of HCl?

The molar mass of HCl, or hydrochloric acid, is 36.46 g/mol. This means that for every one mole of HCl, there are 36.46 grams of the compound.

3. Why is it important to calculate the concentration of a solution?

Calculating the concentration of a solution is important because it allows you to determine the amount of solute present in a given amount of solvent. This information is crucial for many scientific experiments and processes, as it helps to ensure accuracy and consistency in results.

4. What is the difference between normality and molarity?

Normality (N) and molarity (M) are both units of concentration, but they measure different properties of a solution. Normality measures the number of equivalents of a solute per liter of solution, while molarity measures the number of moles of a solute per liter of solution.

5. Can the concentration of a solution change after removing a certain amount of solute?

Yes, the concentration of a solution can change after the removal of a certain amount of solute. This is because the concentration is dependent on the amount of solute and solvent present in the solution. Removing solute will decrease the concentration, while adding more solvent will increase the concentration.

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