[ssstrong]

Hi

A Stock solution of 50% NaOH is available . How would 2L of 10% solution be prepared ? Atomic weight of (Na=23,O=16,H=1)

Some people answed me but without using the atomic weight, so what is the advange of them ??

Best reagrds

Saman SarKo

[Yggdrasil]

This is a dilution problem; taking a solution of one concentration and making a solution of a higher concentration.  Given that you know the concentration of the original solution, you don't need to know the molecular weight to solve the problem.

There are two approaches you can take to calculate the volume of 50% NaOH needed for your 10% solution:

1) Calculate a dilution factor.  10%/50% = 1/5, so you are performing a 5x dilution.  This means the volume of 50% NaOH that you use for making your 10% solution is 2L * (1/5) = 0.4L.

2)  Use the equation c_initialv_initial = c_finalv_final, where c represents the concentrations and v represents the volumes.  This approach is more general and is helpful when the dilution factor is not an easy number to work with (e.g. making a 12.25% solution of NaOH), but mathematically, the two approaches are equivalent.

[Borek]

Ygg: in both cases you are neglecting density changes. They'll be huge in the case of these solutions.

[ssstrong]

Thnx dude ...

this question i found it in of the Proff. Lecture ...

May There is another request in the Question which required using the molecular weight ??

Thnx AgAin ..