The density of a 2.10 M solution of LiBr in acetonitrile (CH3CN) is 0.826 g/mL. What is the molality (m) of this solution? Note: Assume 1000.0 mL of solution for your calculation. Show your work in the space below and report your answer using the correct units and significant figures.
To find out molality first calculate moles of solute i.e LiBr and mass of solvent i.e.acetonitrile
Given ,Molarity of solution =2.10M, volume=1000ml=1L
Molarity=moles of solute/volume of solution in L
2.10mol/L=moles of LiBr÷1L
therefore moles of LiBr=2.10mole
Molar mass of LiBr=86.845g/mol
mass of LiBr=moles×molar mass=
2.10mol×86.845g/mol=182.37 g
Now given density of solution=0.826g/ml
density=mass /volume
0.826g/ml=mass of solution/1000ml
mass of solution=0.826 g/ml ×1000ml=826 g
mass of solution =mass of LiBr+mass of acetonitrile
therefore mass of acetonitrile(solvent)=826g-182.37g=643.63 g=0.64363kg
Now, molality ,m =moles of solute÷kg of solvent
therefore m=2.10÷0.64363=3.26mole/kg
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