[Syaoran]

I need help in finding the grams of LiCl in 275g in a 15% aqueous solution.
Can you help me out with the procedure? I just need a little *Ignore me, I am impatient* in how to start it

Also another question, when i have another question about Mixtures, should i post the problem on the same thread or make a new one?

Thank you Very Much

[Rabn]

First off, let's clarify what it is the question asks.  The words are very important in both your understanding and our ability to help you.  I'm going to assume that you need to find the mass(g) of LiCl in 275g of a 15% solution.  There is a very important distinction between the solution being 15% by weight or volume.  Once you determine if the solution is "by weight" or "by volume" let us know what the difference is. The goal of this forum is to help you understand and enjoy chemistry by guiding you toward the answer.  We have no idea what you know so we start with the basics.

You can start as many threads as you want. However, when you post a question make sure you include any work you have done or how you plan on attempting to solve the problem. It's your job to solve the problem, it's our job help get you there.

[Syaoran]

This was the Original Question
How many grams of CiCl are in 275g of a 15% aqueous solution of LiCl

Also, i was thinking about trying to find the Molarity but i have no idea where does the 15% comes into play. I think i need to make a conversion factor out of the data but i have no idea how.

For some reason, i feel like i dont quite understand the question

[Rabn]

Alright, let's start with an X% solution by volume.

- Say you want a 3% NaCl aqueous solution b/v(this stands for "by volume")
  to make it you would choose a really convenient final volume...like 100mL:
  3% of 100mL is 3 mL, so you would put 97 mL of water in a graduated cylinder(not what you'd really
  use but it's convenient); then you would add NaCl until the meniscus read 100 mL, a change of 3 mL.

Now an X% solution by weight.

- If you instead wanted a 3% aqueous solution b/w(by weight), you would again choose a really   
  convenient final mass for the solution, like 100g.  3% of 100g is 3g. So you would weigh out 97g of
  water into a beaker, then weigh out 3g of NaCl and add it to the water, giving you 100g of solution.


Think about the difference between the amount of NaCl you would need to add to each solution.

That should help clear things up.

[Borek]

My bet is that they are asking about w/w. Call it intuition.

Now an X% solution by weight.

- If you instead wanted a 3% aqueous solution b/w(by weight), you would again choose a really  
  convenient final mass for the solution, like 100g.  3% of 100g is 3g. So you would weigh out 97g of
  water into a beaker, then weigh out 3g of NaCl and add it to the water, giving you 100g of solution.

Finally, if you want w/v solution, you will weight 3 g of NaCl, dissolve it in a small amount of water and fill it up to 100 mL (you may use the same graduating cylinder for that).

That should help clear things up.

Quite contrary I am afraid.