Question

Experts...The density of 1m solution of Nacl is 1.0585 g/mL. The molality of the solution is ?

Answer 1

Hey mate ^_^

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Answer:
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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

#Be Brainly❤️

Answer 2

Explanation:

We know that,

$\sf d = M\bigg( \dfrac{Mw_2}{1000}+\dfrac{1}{m} \bigg)$

$\sf 1.0585 = \bigg( \dfrac{58.5}{1000}+\dfrac{1}{m} \bigg)$

$\sf \dfrac{1}{m} = 1.0585-0.0585$

$\sf \dfrac{1}{m} =1$

$\sf m = 1$

Hence, the molality of the solution is 1 m.