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

Answer:

Molality = 6.0 m

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 : $Molarity ( \frac{1}{molality } +\frac{mol.wt}{1000} )$

∴

$1.023 = 3.23 (\frac{1}{molality } +\frac{150.087}{1000} )$

$\frac{1}{molality } =( \frac{1.023}{3.23} - \frac{150.087}{1000} )$

$\frac{1}{molality } =0.3167 - 0.1500$

$\frac{1}{molality } = 0.1667$

Molality  = $\frac{1}{0.1667}$

Molality = 5.999 m

Molality ≅ 6.0 m