Question

the molality of 0.05 M HCL(aq.) in 5L solution at 4 degree C is

Answer 1

Answer:

Let the volume of solution be 1 litre∴ Molarity=

Volumeofsolution(inL)

MolesofNaOH

∴0.05=Moles of NaOH

Now, Mass of solvent=Volume×Density

=1×1

=1kg

∴ Molarity=

Massofsolvent(i.e.H

2

O)inkg

MolesofNaOH

=

1

0.05

=0.05m

Answer 2

We have to find the molarity of 0.05 M HCl in 5 L aqueous solution at 4°C.

solution : molarity of solution is 0.05 M, it means 0.05 mole of HCl is present in 1L of solution.

∴ no of moles of HCl in 5L solution = 0.05 × 5 = 0.25 mole

mass of HCl = no of moles × molar mass

= 0.25 × 36.5

= 9.125 g

at 4°C , density of water (because solution is aqueous so solvent must be water) = 1 g/cm³

mass of solution = volume × density

= 5L × 1 g/cm³

= 5000 ml × 1g/cm³

= 5000 g

mass of solvent = mass of solution - mass of solute

= 5000g - 9.125 g

= 4990.875 g

now molality = no of moles of solute/mass of solvent in kg

= 0.25 /(4990.875/1000)

= 250/4990.875

= 0.0500914168 molal ≈ 0.05 molal

Therefore molality of solution is 0.05 molal.