Question

The molality of 1M solution of NaCl (specific gravity 1.0585g/ml) is :

Answer 1

Thanks for the question!

It is definitely a very interesting question to solve and do some brainstorming.

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Take a hypothetical sample of 1.000 L of the solution.
(1.000 L) x (1 mol/L) x (58.4430 g NaCl/mol) = 58.4430 g NaCl

(1000 mL) x (1.0585 g/mL) = 1058.5 g total

(1058.5 g - 58.4430 g NaCl) = 1000.057 g H2O

(1.000 L) x (1 mol/L) = 1.000 mol NaCl

(1.000 mol) / (1000.057 g / 1000 g) =

(1.00000 mol) / ((1 000.057 g) / (1000 g)) = 1.000 mol/1000 g = 1.0 m
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Hope it helps and solves your query!!

Answer 2

Let us consider the volume of solution to be 1L.
The total weight of the solution would be 1.0585 g/mL×1000mL = 1058.5g
1M solution means that 1 mole of solute (NaCl) would be present in 1000mL. Hence, the weight of NaCl in the solution is 23+35.5 = 58.5g. The weight of solvent would be equal to the weight of solution - the weight of the solute. Hence, the weight of the solvent = 1058.5-58.5 = 1000g
Molality of the solution is defined as the number of moles of solute present in 1000g of solvent. As there is just one mole of NaCl present in the solution (which has 1000g of solvent), the molality of the solution would be 1.