[trishasales714]

If I had 5 g of a solid with molecular weight 53.0 g/mol in 1 L of water, wouldn't I just convert grams to moles by 5 g / 53 g/mol, then dividing this value by 1 L to get the molarity? I'm just not sure if this is right since molarity is moles of solute / liters of SOLUTION not solvent.

[AWK]

moles in 1L equals to molarity

[trishasales714]

moles in 1L equals to molarity

Yes. Isn't it liters of solution and not solvent? So I can't just calculate the moles of solid and divide that by 1 L of water (solvent). I don't know how to take the solid into account since it's the volume of the solution.

[AWK]

5 g in 1L  may change volume ± 2 ml which is ±0.1-0.2 %. For most cases we neglect such error.

[Borek]

Isn't it liters of solution and not solvent? So I can't just calculate the moles of solid and divide that by 1 L of water (solvent). I don't know how to take the solid into account since it's the volume of the solution.

You are right, but when the amount of solid is small compared with the amount of solvent (and 5 g against 1000 g definitely is), we can safely ignore change in volume.

Even for 50 g of NaCl the error made assuming no change in volume won't be large, but if you happen to need a very precise result you can either use density tables for calculations, or if all you have to is to prepare a solution, dissolve your solid in a lower amount of water and then fill up to 1 L.

[trishasales714]

For something like 0.5 g in 100 g then change in volume due to the solid would be negligible as well right?

Isn't it liters of solution and not solvent? So I can't just calculate the moles of solid and divide that by 1 L of water (solvent). I don't know how to take the solid into account since it's the volume of the solution.

You are right, but when the amount of solid is small compared with the amount of solvent (and 5 g against 1000 g definitely is), we can safely ignore change in volume.

Even for 50 g of NaCl the error made assuming no change in volume won't be large, but if you happen to need a very precise result you can either use density tables for calculations, or if all you have to is to prepare a solution, dissolve your solid in a lower amount of water and then fill up to 1 L.

[Borek]

For something like 0.5 g in 100 g then change in volume due to the solid would be negligible as well right?

Exactly the same as 5 g in 1 L, it is the ratio that matters and it is identical.