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What mass of cesium bromide must be added to 0.500 L of water (d = 1.00 g/mL) to produce a 0.400 m solution? What are the mole fraction and the mass percent of CsBr?

Short Answer

Expert verified

The number of moles of Cesium bromide having molality 0.400m is 0.2moles. The mole-fraction of 0.2mol of cesium bromide is 0.0071.The mass percentage of the 42.6g of CsBr in 500g of water, solvent is 7.9%.

Step by step solution

01

Concentration of the Solution

Solution can be formed from the dissolution of the solute in the solvent. The solute decides the nature of the solution.

Molality can be defined as the ratio of the mass of the solute to the mass of the solution present in kilogram measure.

Molality=Number of MolesMass of the Solution ( in kilogram )

Number of moles can be defined as the ratio of the mass of the atom/molecule and molar mass of the atom/molecule.

Number of Moles=MassMolar Mass

Mole Fraction can be defined as the ratio of the number of moles of the solute to the number of moles of solvent and number of solute.

Mole Fraction=Number of Moles of One ComponentNumber ofMoles of the Solvent+Number of Moles of Solute

Mass%=Mass of soluteMass of solution×100

Density can be defined as the ratio of the mass of the matter to the volume of the matter.

Density=MassVolume

02

Calculation

Volume of Water=0.500L=500mL

Density of the Water =1.0g/mL

Density of Benzene=MassVolume1.0g/mL=Mass2.66mLMass=1.0g/mL×500mLMass=500g

Mass of Solvent, Water =500g=0.500kg

Molality of solute, CsBr =0.4m

Molality=Number of MolesMass of the Solvent ( in kg )0.4m=Number of Moles0.500kgNumber of Moles=0.2

Number of moles of water =0.2m

Number of Moles=MassMolar Mass=500g18g/mole=28mole

Number of moles of CeBr =0.2moles

Number of moles of water =28moles

role="math" localid="1659355972498" Mole Fraction=Number of Moles of One ComponentNumber of Moles of the Solvent+Number of Moles of SoluteMole Fraction of CsBr=0.20.2+28Mole Fraction of CsBr=0.228.2Mole Fraction of CsBr=0.0071

Number of Moles=MassMolar MassNumberofMoles of CsBr=Mass213g/mol0.2mole=Mass213g/molMass of CsBr=0.2mol×213g/molMass of CsBr=42.6g

Mass of CsBr =42.6g

Mass of water =500g

 Mass%=Mass of SoluteMass of the Solution×100=42.6g500+42.6g×100=42.6g542.6g×100=7.9%

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