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.
"Density = \\frac{mass}{volume}"
"Mass= density \u00d7 volume"
"Mass= 1.09\u00d71000= 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\u00d7100)}{1682.82}=0.0802M"
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