Question

A solution contains 25% glucose in water calculate molality and mole fraction of is component (density of the solution is 1.2gml-1)​

Answer 1

molality

let the mass of solution be 100gm

according to question

mass of glucose = 25% of 100gm

= 25gm

mass of solvent = 100gm - 25gm= 75gm

density of the solution = 1.2 gm/l

mass of solution/volume of solution=1.2mg/l

volume of solution=massof solution/density

= 100gm/12gm/l

= 83.33l

molar mass of glucose = 180gm/mol

given mass of glucose = 25g

number of moles of glucose= given mass/ molar mass

=25gm/180gm/mol

= 0.139 mol

m(molality) = number of moles of solute/ mass of solvent in kg

= 0.139 / 75x10^-3kg

= 185m

mole fraction

mole of solute = 0.139 mol

mole of solvent=given mass of water/ molar mass of water

= 75g/18g/mol

= 4.17 mol

mole of solution= mole of solute + mole of solvent

= 0.139+4.17=4.31 mol

mole fraction of solute = mole of solute/ mole of solution

= 0.139 / 4.31

= 0.032

mole fraction of solvent = moleof solvent /mole of solution

= 4.17 / 4.31

= 0.97

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