[mac227]

Determine the amounts of solute and solvent needed to prepare:

1) 2 M aqueous NaCl solution

2) 1 m aqueous KCl solution

3) 25% by mass aqueous MgCl2 solution

My Answers:
1) 116.88g NaCl and 1 L H2O

2)  7.4 g KCl and 100 mL H2O

3) 95.211g MgCl2 and 380.844 g H2O

Did I screw up anywhere?

[Nobby]

Number 3 I would say to make 1kg you need 250 g MgCl₂ and 750 g water. The volumen is given if the specific gravity is also known.

[sjb]

Determine the amounts of solute and solvent needed to prepare:

1) 2 M aqueous NaCl solution

2) 1 m aqueous KCl solution

3) 25% by mass aqueous MgCl2 solution

My Answers:
1) 116.88g NaCl and 1 L H2O

2)  7.4 g KCl and 100 mL H2O

3) 95.211g MgCl2 and 380.844 g H2O

Did I screw up anywhere?

Check your definitions, I think these are all not quite correct.

[Borek]

1) 2 M aqueous NaCl solution

I guess 1L of the solution?

2) 1 m aqueous KCl solution

m or M? This can be a serious difference (molality vs molarity). And - again - how much?

3) 25% by mass aqueous MgCl2 solution

How much?

1) 116.88g NaCl and 1 L H2O

If you plan on making 1L, that's wrong. This way you will get more than 1L. Correct approach is to get solid, add some water, dissolve it and then fill up to 1L.

3) 95.211g MgCl2 and 380.844 g H2O

That's 20%, not 25%. And rather strange amount (476.055g).

[mac227]

All my directions say is "determine the amounts of solute and solvent needed to prepare the following solutions"

1) 2 M means 2 moles per liter right?   So my answer should be correct

2) Yes it is small m for molality which is moles solute per kg solvent.  THe directions say perpare 1 m (molality) aqueous KCl solution.  (I am confused by how to do this)


Water is the solvent and 1g=1 mL ??  So 1000ml=1kg.     molecular weight of KCl is 74.555

So 74.555 g of KCl = 1 mole of KCl           and to get 1 m the moles of solute and kg solvent have to be the same number right?

So 1m= 5 moles KCl/5kg H2O??       I'm confused, it definately feels like im doing something wrong



3)  % mass =  (mass solute/mass solution) x 100
    so 25% = 95.211/380.844 x 100

[Nobby]

First it is not given how much volumen or mass you have to prepare. I guess 1 l or 1 kg.

1. 2 mol NaCl = 2 x 58,5 g = 117 g

This you add in a 1 l measured flask and add water until reach 1 liter.

2. 1 mol KCl = 74,5 g

This put in a beaker glass what is already zero balanced on an balance.  You add water until you reach 1 kg, should be 925,5 g

3. I explained already.  25% w.w means 25 % of the mass of the solution is the amount of salt

So you take 250 g salt and 750 g water to make 1 kg.

[Borek]

1) 2 M means 2 moles per liter right?   So my answer should be correct

No, its is not. If you mix 1L of water with 116.88g NaCl final volume will be 1.04L, not 1L.

So 1m= 5 moles KCl/5kg H2O??       I'm confused, it definately feels like im doing something wrong

That's correct. No idea why 5 moles per 5 kg and not 1 mole per 1 kg, but you are right. Now just convert moles to grams.

3)  % mass =  (mass solute/mass solution) x 100
    so 25% = 95.211/380.844 x 100

Number are OK (even if I have no idea why you selected these specific ones, instead of some simpler ones, like 100/400). However, mass of the SOLUTION is sum of masses of SOLUTE and SOLVENT. You need to take it into account.

[Borek]

2. 1 mol KCl = 74,5 g

This put in a beaker glass what is already zero balanced on an balance.  You add water until you reach 1 kg, should be 925,5 g

It is obvious you misunderstood molality definition, read it again.

[mac227]

molality=moles solute/kg solvent

so a 1 m solution is 1 mol solute / 1 kg solution.   

So one mole KCl + 1 kg water = 1 m aqueous KCl solution

As for the first problem (number 1) I understand that if I add 2 moles of NaCl to a 1 liter solution I will increase the volume of that solution.  So how do I go about accounting for this difference in volume?

[Borek]

molality=moles solute/kg solvent

so a 1 m solution is 1 mol solute / 1 kg solution.   

So one mole KCl + 1 kg water = 1 m aqueous KCl solution

Yes.

As for the first problem (number 1) I understand that if I add 2 moles of NaCl to a 1 liter solution I will increase the volume of that solution.  So how do I go about accounting for this difference in volume?

I already gave you an answer in an earlier post.