Calculate osmolality of a 0.1% and a 0.9% NaCl.?

[Wek]

I tried to calculate the osmolality (mOsm/kg) of a 0.1% and a 0.9% NaCl solution but I am not sure I'm doing this right.

First I calculated the NaCl grams in a liter of solution, so for 0.1% I have 0.01g of NaCl. Then I divided by the 58.44 (MW) to get moles. Then I divided the moles by 1 liter (since the osmolality is per 1 kg of solution) to get the molality. Then I multiplied the molality by the dissociation constant of NaCl (1.8). Then I divided by 1000 to get mOsm.

The final answer I got is 0.308008214 mOsm/kg. Can someone double check my math?

Thanks

 

[Borek]

First I calculated the NaCl grams in a liter of solution, so for 0.1% I have 0.01g of NaCl.

No, 0.01g is wrong.

In general your approach looks OK to me, with one exception. Where did you get the 1.8 for a "dissociation constant" from?

 

[Wek]

The 1.8 dissociation constant was given.

How much would the the NaCl grams of a 0.1% solution be then?
The way I calculated the NaCl grams was by reasoning that if 1% = 1g/100ml then in 1000ml there would be 10g (10g/1000ml = 1%). So for a 1 liter of 0.1% NaCl solution there would be 10g*(0.1/100) = 0.01g.

 

[Borek]

if 1% = 1g/100ml then in 1000ml there would be 10g (10g/1000ml = 1%).

Correct, but

So for a 1 liter of 0.1% NaCl solution there would be 10g*(0.1/100) = 0.01g.

this is wrong and I can't even understand what you did.

Try to directly apply percentage definition:

\frac{\text{mass of NaCl}}{\text{mass of 1L of solution}} 100\% = 0.1\%

 

[Wek]

Hm. If I use that formula I would need to use 0.001g of NaCl per 1kg of water to get a 0.1% solution.

(0.001g/1kg)*(100) = 0.1%

So to make 1 liter of 0.1% NaCl I need 0.001g of NaCl?

 

[Borek]

(0.001g/1kg)*(100) = 0.1%

You divide grams per kilograms and you think it is OK?

Using your approach if you have 1 cent and I have 1 dollar we have both the same amount of money, as 1=1. You have to watch your units, always.