Question

An aqueous solution is 3.23M in tartaric acid (C4H06). The solution's density is 1.023 g/mL.

Calculate the solution's molality in tartaric acid.

Answer 1

Molality = 6.0 m

Step-by-step explanation:

The molecular weight of tartaric acid = 150.087 g/mol

Given that:

Density of the solution = 1.023 g/mL

Molarity = 3.23M

Density is given as :
`Molarity ( (1)/(molality ) +(mol.wt)/(1000) )`

∴

`1.023 = 3.23 ((1)/(molality ) +(150.087)/(1000) )`

`(1)/(molality ) =( (1.023)/(3.23) - (150.087)/(1000) )`

`(1)/(molality ) =0.3167 - 0.1500`

`(1)/(molality ) = 0.1667`

Molality =
`(1)/(0.1667)`

Molality = 5.999 m

Molality ≅ 6.0 m