how can i find out the density of 5M NaOH
If you are in a course, you will learn this. Chemistry is a laboratory course. Density is a ratio of mass to volume. Do you have a 5 M solution of NaOH?
Otherwise, look in a handbook.
Work out how many moles of NaOH in a litre,
Work out the mass of 1 mole of NaOH, add to the mass of 1litre of water.
Thats the density (approximately - doesn't take into account the volume of the added NaOH)
In general - this is wrong. With such approximations you never know what the error is - and it can be everything. For ammonia you will estimate that 5M solution is around 1.085 g/mL, while in reality it is 0.963 g/mL. For sulfuric acid your estimate will be 1.49 g/mL - compare it to the real value of 1.29 g/mL. That means -11% and +16% errors.
The only correct approach is to use density tables from some handbook, like CRC handbook for example. There are short density tables on my site. Note, that in many hanbooks density tables contain only densities of solutions of given % w/w concentration, so you have to convert between M and %. That's not a simple one step operation, as to convert you have to know density first. So, you either do it iteratively (assume some density, convert, check new density in tables, use it to convert and so on, depending on the accuracy you need one or two iterations will be most likely enough) or you prepare M vs density table (that makes sense if you will be doing conversions often) or you look for some concentration calculator with built in density tables, that does conversion for you automatically
I would suggest a simple weighing trial however the basic solution may scratch the glassware ; simply measure out a certain volume of the solution and then weigh it however be sure to tare the measuring apparatus.
It is 1.182 g/mL at 25oC according to Sigma Aldrich.
Let's see what the error would be assuming no DV of mixing.
The density of solid NaOH is 1.829 (according to Wiki). We need 5*40.00 g for 5 moles which will be diluted with sufficient water to form 1 L. Assuming no DV of mixing, we will add 200.0 g or 109.4 mL (200.0 g/1.829 g/cm^3) to 890.6 mL of water (1000 mL - 109.4mL). This will give us the 1L we need and will weigh 1090.6 g for a density of 1.091. The relative percent mixing error is 100* (1.091 g/mL - 1.182 g/mL) / 1.182 g/mL = -7.699%
That's a little too much error for my liking.
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