Dear Kamyar,
At 20 °C to prepare saturated salt solution which its density is 1.2 g cm-3, 357 grams of NaCl is added to one liter of water (concentration of 26.3%).
Density of saturated solution required is 1.2 g/ml (1 cm3 = 1 ml).
Means 1 ml Solution requires 1.2 g NaCl
So, for 600 ml solution,
1.2 g x 600 ml = 720 g NaCl required.
Use this..its that simple...!!
http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html
Dear Hitesh Chavda Dissolving 720 grams of salt in 600 ml of water, appears rather impractical. Although my initial calculations led to the same result.Thank you for taking the time to answer the question.
Mina Karimi-Avargani Thank you for your answer. Could you please provide the related calculations?
Mohamed Khedawy Thanks!
Although the links provide useful information regarding molarity, they do not provide any density-based calculations.
(120 g NaCl + 600 g water) / (600 ml water)
= 720 g / 600 ml
= 1.2 g/ml.
Please note saturated solution means maximum amount of solute gets dissolved in solvent.
In line to that 357mg/ml at 25 C, is water solubility of NaCL.
Hence 214.2 g of NaCl in 600 ml water at 25 C will produce Saturated solution of NaCL with density of 1.357 g/ml
The (molar) solubility of the salt (S) for a pure NaCl aq. sol. (mol / dm3) can be obtained as I have shown elsewhere at this forum: https://www.researchgate.net/post/What_is_the_maximum_concentration_of_sodium_chloride_in_water
Density of saturated solution required is 1.2g/cm^3 = 1.2 g/ml
So, for 600 ml solution,
Therefore, the total NaCl required would be: (1ml/600 ml) x 1.2 g = 720 g
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