Conversions (1 Viewer)

[leehuan]

(a) Calculate the molarity of a solution prepared by dissolving 0.345 g calcium chloride in

enough water and diluting to 250.00 mL. (2 marks)

The answer to this part is 0.0124 mol L^-1 at 3 s. f. Just use the rounded answer for convenience please.

(b) What is the concentration of the chloride ion in (a) in %(w/v)? (2 marks)

Ok, so I easily understand that [Cl^-] = 0.0248 mol L^-1.
To my understanding, we need to find the mass of the chloride ions present to use the formula %(w/v) = mass of solute/volume of solution * 100%

But the answers just did this.



But I don't get how this works. 0.0248 times molar mass of chlorine just converts the concentration to g L^-1. That per-litre is throwing me off big time. Please explain what I missed.

 

[turntaker]

Why are you doing chem...

 

[InteGrand]

(a) Calculate the molarity of a solution prepared by dissolving 0.345 g calcium chloride in

enough water and diluting to 250.00 mL. (2 marks)

The answer to this part is 0.0124 mol L^-1 at 3 s. f. Just use the rounded answer for convenience please.

(b) What is the concentration of the chloride ion in (a) in %(w/v)? (2 marks)

Ok, so I easily understand that [Cl^-] = 0.0248 mol L^-1.
To my understanding, we need to find the mass of the chloride ions present to use the formula %(w/v) = mass of solute/volume of solution * 100%

But the answers just did this.



But I don't get how this works. 0.0248 times molar mass of chlorine just converts the concentration to g L^-1. That per-litre is throwing me off big time. Please explain what I missed.

Are the answers from a source you expect to be reliable?

 

[leehuan]

Are the answers from a source you expect to be reliable?

It's a James Ruse prelim exam question

 

[InteGrand]

(a) Calculate the molarity of a solution prepared by dissolving 0.345 g calcium chloride in

enough water and diluting to 250.00 mL. (2 marks)

The answer to this part is 0.0124 mol L^-1 at 3 s. f. Just use the rounded answer for convenience please.

(b) What is the concentration of the chloride ion in (a) in %(w/v)? (2 marks)

Ok, so I easily understand that [Cl^-] = 0.0248 mol L^-1.
To my understanding, we need to find the mass of the chloride ions present to use the formula %(w/v) = mass of solute/volume of solution * 100%

But the answers just did this.



But I don't get how this works. 0.0248 times molar mass of chlorine just converts the concentration to g L^-1. That per-litre is throwing me off big time. Please explain what I missed.

Sources online seem to confirm your idea of how to find %(w/v), so you should probably just find the mass of chloride ion and use the formula you gave. To do it the answers' method, they should have found the mass of chloride ion in 250 mL (i.e. 0.25 L) by doing the calculation as:

(0.0248 mol/L) * (35.45 g/mol) * (0.25 L).

So they left out a 0.25 L I think. So the answer should probably be 0.25 of what they got.