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The density of a 2.10 M solution of LiBr in acetonitrile (CH3CN) is 0.826 g/mL. What...

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.

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Answer #1

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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