1.

Calculate the molarity and mole fraction of the solute in aqueous solution containing `3.0 g` of urea per `250gm` of water (Mol. Wt. of urea = 60)A. ` 0.2m ,0.00357 `B. ` 0.4m, 0.00357 `C. ` 0.5m, 0.00357 `D. ` 0.7m, 0.00357 `

Answer» Correct Answer - A
Wt.of solute(urea)dissolved=`3.0g`
wt. of the solvent (water)=`250g`
Mol.wt.of the solute=`60`
`3.0gm`of the solute =`(3.0)/(60)mol es `
`=0.05Mol es`
thus `250g`of the solvent contain =`0.05`Moles of solute
`:.1000g`of the solvent contain =`(0.05xx1000)/(250)=0.2`moles
Hence molality of the solution =`0.2m`
In short,
Molality = No . of mol es of solute /1000g of solvent
`:.` Molality `=(3//60)/(250)xx1000=0.2m`
calculation of the mole fraction`3.0 g` of solute `=3//60 "moles" =0.05 "mole"`
`250g`of water `=(250)/(18)`moles =`13.94`moles
`:.` Mole fraction of the solute `=(0.05)/(0.05+13.94)=(0.05)/(13.99)=0.00357`


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