Question

calculate the molality solution containing 1.5mole in 675ml of solvent given density 1gml​

Answer 1

Answer:

1)Only if we consider the solvent in this question is water ,we can consider the mass of solvent to be 675mL =675 g

since for water:

1L =1 kg

And thus

Molality of solution = Moles of solute / kg of solvent

=1.5/0.675 = 2.2 m

2)

No. of moles of Ba(NO3)2 = Mass/Molar mass =750/261.32 = 2.8700 moles

Density = 1g mL-1

So, Volume = 374 mL (as density = mass/ volume)

Molarity = No. of moles / Volume of solution (L)

Molarity = 2.87moles374mL×1000mL1L

Molarity = 7.67 M