Question

Q ) 200 ml of 1M HCI is mixed with 800 mL of 2 M HCl. Find molality of the final solution. Given, d=1.2 gm/mL​

Answer 1

The molality of the final solution is 1.59 m.

Step-by-step-explanation:

$m=\frac{n_{HCL} }{W_{solvent} }$

M. moles of HCl = (200 X1) + (800×2) = 1800

Moles of HCL=1800/1000=1.8 moles

Wsolvent= $W_{solvent} - W_{solute}$

$w_{sol}$ =dxv=1.2x1000=1200gm

$W_{solute}$ = (1.8 X36.5) gm=65.7gm

m= $[\frac{1.8}{1200-65.7}] x1000=1.5868$≅ 1.59m

What is meant by molality:

The quantity of a material dissolved in a specific mass of solvent is known as its molality (m), also known as molal concentration. The moles of a solute per kilogram of a solvent is how it is defined.

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