# DE Problem

#### kanki

Hi I'm a newbie and i have a problem here that i cannot solve, or i don't know how to solve. Hope you experts can help me out!
Here it is:
A tank contains 10 kg of sodium hydroxide in 1000 litres of water. Water is continuosly added to the tank at a rate of 5 litres/min so that the mixture is diluted evenly. At the same time, the solution flows out at the same rate. Initially, there were 10 grams of sodium hydroxide in every litre of water. How long will it take for the concentration of the solution to drop to 3.2 grams per litre?

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#### saltydog

Homework Helper
kanki said:
Hi I'm a newbie and i have a problem here that i cannot solve, or i don't know how to solve. Hope you experts can help me out!
Here it is:
A tank contains 10 kg of sodium hydroxide in 1000 litres of water. Water is continuosly added to the tank at a rate of 5 litres/min so that the mixture is diluted evenly. At the same time, the solution flows out at the same rate. Initially, there were 10 grams of sodium hydroxide in every litre of water. How long will it take for the concentration of the solution to drop to 3.2 grams per litre?
Hello Kanki. You know, you need to show a little work dude. It's best for you that way don't you think? Just make some attempt at solving it, anything, just move the numbers around the paper randomly, whatever, and show us what you did and explain what you think should be done. Here's one we worked on earlier (click on the link below). I bet a dollar if you follow it closely, you can figure out yours. Try it, then come back and say, "look, I tried to do this one, here's what I did <insert something here, anything>. Can you guys explain where I'm going wrong?" Or someone may just solve it. Well, what I said is what the general policy is anyway.

EvLer's mixing problem

Welcome to PF! Last edited:

#### kanki

Oh i'm so excited now!
I have solved the problem after refering to the thread you gave me just now!
It's awesome... wow... and this problem is much more easier than EvLer's.
Thanks!!!

Here's the solution:

The rate of change of volume is constant. so V is a constant. And i just have to consider the change in the mass of the salt.

V = 1000 litres

let s = mass of salt in the solution

$$s(0)=10kg$$
$$ds/dt=-5(s/v)$$ (after reading through the thread, i realise that i missed that mass of salt is volume x concentration)

$$ln|s|=-5/1000(t)+ln10$$

concentration is 3.2 grams/litres and it's 3.2kg in 1000 litres.

By substituing the value s=3.2 kg
finally,

t = 227.89 minutes = 3 hours 48 seconds

#### saltydog

Homework Helper
kanki said:
Oh i'm so excited now!
I have solved the problem after refering to the thread you gave me just now!
It's awesome... wow... and this problem is much more easier than EvLer's.
Thanks!!!

Here's the solution:

The rate of change of volume is constant. so V is a constant. And i just have to consider the change in the mass of the salt.

V = 1000 litres

let s = mass of salt in the solution

$$s(0)=10kg$$
$$ds/dt=-5(s/v)$$ (after reading through the thread, i realise that i missed that mass of salt is volume x concentration)

$$ln|s|=-5/1000(t)+ln10$$

concentration is 3.2 grams/litres and it's 3.2kg in 1000 litres.

By substituing the value s=3.2 kg
finally,

t = 227.89 minutes = 3 hours 48 seconds
That's 3 hours and 48 minutes right?

Last edited:

#### kanki

Oh thanks for correcting me, i'm sorry! ha ha...

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