Question

Lithium bromide as a aqueous solution with molar mass of 86.8g/mole and molality of 

1.35m and a density of 1.09g/mL. Calculate its molarity.

Answer 1

$Density = \frac{mass}{volume}$

$Mass= density × volume$

$Mass= 1.09×1000= 1090g$

Mass of solution = mass of solute + mass of solvent

Mass of solution = (no. of moles of solute) × (molar mass of solute )+ mass of solvent

1800 = (1.35 × 86.8)+ mass of solvent

Mass of solvent = 1800 - 117.18 = 1682.82 g

$Molarity = \frac{(1.35×100)}{1682.82}=0.0802M$