Creating and using an equation

[Hundra]

Homework Statement

Water from the middle of an ocean contains about 1% salt. How much salt do you have to add per liter in order for the salt-level to reach 34%?

Homework Equations

I can't manage to put together the equation.

The Attempt at a Solution

I know the answer is supposed to be 0,5 kg. Somehow end up with 2 kg.

Thanks a lot for any input and help :)

 

[berkeman]

Homework Statement

Water from the middle of an ocean contains about 1% salt. How much salt do you have to add per liter in order for the salt-level to reach 34%?

Homework Equations

I can't manage to put together the equation.

The Attempt at a Solution

I know the answer is supposed to be 0,5 kg. Somehow I end up with 2 kg.

Thanks a lot for any input and help :)

Is that % by weight? How much does salt weigh?

 

[Hundra]

The problem itself does not specify anything concerning weight. Thats what I am struggling with.

There is an example given earlier that starts with "A saline solution weigh 6kg and contains 5% salt (...)". Can that be used to solve our problem?

 

[tiny-tim]

Welcome to PF

Hi Hundra! Welcome to PF!

I know the answer is supposed to be 0,5 kg. Somehow I end up with 2 kg.

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

 

[Hundra]

Hello, thanks :)

There is nothing (mathematically)wrong with the calculations I am doing. The problem is that I don't know how to produce an equation when I don't have any information regarding how much salt weigh (except from what I wrote in my previous post). I've scribbled a bit on a piece of paper in terms of solving it for another form of measurement than kg, but it doesn't really make any more sense.

 

[berkeman]

Hello, thanks :)

There is nothing (mathematically)wrong with the calculations I am doing. The problem is that I don't know how to produce an equation when I don't have any information regarding how much salt weigh (except from what I wrote in my previous post). I've scribbled a bit on a piece of paper in terms of solving it for another form of measurement than kg, but it doesn't really make any more sense.

I would think that it would be pretty easy to look up the density of salt on the Internet...

 

[Gigasoft]

You don't need to know the density of salt in solid form. You can't just add the volume of a block of salt to the volume of the water to get the volume of the final solution. What you need to know is the density of 1 liter of ocean water. To find this, you must make an assumption about the temperature and the pressure of the solution. The density can then be looked up in a chart, or found using a formula such as the one used on http://www.csgnetwork.com/h2odenscalc.html".

Key questions:
In a liter of the original solution, how much salt is there in it?
If you add X amount of salt, how much salt is there now?
What is the mass of the original solution?
What is the mass of the solution after adding X amount of salt?

 

[Hundra]

This is from a junior high math-book (13-14yo's). Its not supposed to be that difficult. Basically you are supposed to create an equation and solve it.

The only information presented is what I wrote in the first post. In addition there is an example box on the page aswell, where it states, as previously posted:

There is an example given earlier that starts with "A saline solution weigh 6kg and contains 5% salt (...)". Can that be used to solve our problem?

 

[Gigasoft]

No, it can't. But if you assume that one liter of water with 1% salt weighs the same as one liter of fresh water, you get the answer 0.5 kg. All you need to assume is the mass of one liter of ocean water (here, they have equated it to 1 kg), and all the other information you need to write down an equation is given in the problem. The four questions I listed are then easily answered and will provide the four numbers that make up the equation.

 

[Hundra]

I went with this:

Assuming 1 liter ocean water weighs about 1 kg. If 1% is salt, that's 10g, so the other 990g is water. If we want a solution with 34% salt, that means that the rest (=66%) is water. So the 990g is 66%, and then 34% must weigh 510g. If we finally subtract the original 10g, that means we have to add 500g to get there.