Question

Problem
The density of 3 M solution of Naci
1.25 g ml. Calculate the molality of the
solution​

Answer 1

Answer:

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Explanation;

2.08

Answer 2

Given :

Density of solution = 1.25 g / ml

M = 3 mol / lit

Molar Mass of NaCl = 58.5

We know that ,

$number \: of \: moles = \frac{weight}{molecular \: weight}$

Weight = No. of moles × Molecular weight

Mass Nacl in 1 litres solution = 3 × 58.5

= 175.5 g

$density = \frac{mass}{volume}$

Mass = density × volume

Mass of 1 litre solution = 1000 × 1.25

= 1250 g

Mass of water in solution = Mass of solution - Mass of solute

= 1250 - 175.5

= 1074.5 g

Mass of water in solution = 1.074 kg

$molality = \frac{no. \: of \: moles \: in \: solute}{mass \: of \: solvent \: in \: kg}$

$\frac{3}{1.074}$

Molality = 2.79 m

Molality of the solution is 2.79 m