What Is the Original Concentration of the Solution?

AI Thread Summary
To find the original concentration of a solution, the problem states that diluting 500 ml to 2 liters results in a concentration of 20 g/L. The calculations show that the total mass of solute in the diluted solution is 40 g. Since the amount of solute remains constant during dilution, the original concentration in the 500 ml solution is determined to be 40 g/500 ml. This simplifies to an original concentration of 80 g/L. The discussion confirms that the original concentration can be accurately calculated based on the dilution principle.
DB
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Having a little trouble tackling this one.
"What is the concentration of a 500 ml solution if, when diluted to a volume of 2 Liters, its new concentration becomes 20 g/L?"

Here what I did, all in ml.

\frac{x}{500+1500}

\frac{x}{2000}=\frac{20}{1000}

\frac{20*2000}{1000}=40=x

\frac{40}{2000}=\frac{10}{500}

Obviously the anwser can't be 10g/500ml, can someone help me out?
Thanks
 
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DB said:
\frac{x}{500+1500}

\frac{x}{2000}=\frac{20}{1000}

\frac{20*2000}{1000}=40=x
Right. "x" is the quantity of "stuff" in the mix, in g. That stuff doesn't change when you dilute it, so 40g of stuff must have been in the original 500ml. Now can you calculate the original concentration?
 
O so do you mean that solute is 40g for 2000ml and as well for 500ml? Making the answer 40g/500ml?
 
Yes. That's what I would say.
 
Thanks for all your help :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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