Question

The density of 3 M solution of NaCl is 1.25 g ml-1. Calculate molality of the solution​

Answer 1

Answer:Hope this helps you

Explanation:

Answer 2

Molarity is the number of moles present in 1 L of the solution .

Given , molarity = 3 M .

$\mathtt{Molarity=\dfrac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:(in\:litres)}}$

Let the number of moles be n.

$\implies 3mol/l=\dfrac{n}{1l}\\\\\implies n=3mol/l\times 1l\\\\\implies n=3mol$

So number of moles of solute ( NaCl )  is 3 .

$\mathtt{Number\:of\:moles=\dfrac{Mass\:given}{Molecular\:mass}}$

Molecular mass of NaCl = 23 + 35.5

⇒ Molecular mass = 58.5

$\implies 3=\dfrac{m}{58.5}\\\\\implies m=58.5\times 3\\\\\implies m=175.5g$

Mass of NaCl = 175.5 g .

Now it is given that density of the solution is 1.25 g/ml .

Volume = 1 l = 1000 ml .

Mass = density × volume

⇒ mass = 1.25 g/ml × 1000 ml

⇒ mass = 1250 g

This is the mass of solution ↑

Mass of solvent = mass of solution - mass of solvent .

⇒ Mass of solvent = 1250 g - 175.5 g

⇒ Mass of solvent = 1074.5 g

⇒ Mass of solvent = 1.0745 kg .

$\mathtt{Molality=\dfrac{moles\:of\:solute}{mass\:of\:solvent\:inkg}}\\\\\implies \dfrac{3}{1.0745}\\\\\implies 2.7919\\\\\boxed{\approx 2.792}$