The mole fraction of cesium chloride is 0.06296.
The mole fraction is the product of the number of molecules of a specific component in a mixture and its total molecular weight.
By mixing 60.0 g of water with 42.9 g of cesium chloride, a solution is created.
The volume of the solution is given as 63.3 ml.
The solute in this aqueous solution is cesium chloride. From their given masses and known molar masses, we require the moles of water and cesium chloride:
n( CaCl ) = 42.9 g × ( mol ) / ( 168.36 g )
n( CaCl ) = 0.254 mol
n( water ) = 60.0 g × ( mol ) / 18.01 g
n( water ) = 3.78 mol
Therefore, the mole fraction of cesium chloride solute will be:
( 0.254 mol ) / ( 0.254 mol + 3.78 mol ) = 0.06296
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