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A sugar syrup of weight 214.2 g contains 34.2gofsugar(C_(12)H_(22)O_(11)).Calculate(i)molal concentration (ii) mole fraction of sugar in the syrup.

Answer» <html><body><p></p>Solution :Weight of sugar syrup = 214.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> g<br/> Weight of sugar present = 34.2 g <br/> <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass of sugar `(C_(12)H_(22)O_(11)) = 342` <br/> (i) <a href="https://interviewquestions.tuteehub.com/tag/calculation-907729" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATION">CALCULATION</a> of molality : <br/> Number of moles of sugar = `34.2/342 = 0.1` <br/> Weight of water `=214.2 - 34.2 = 180 g = 0.18 g` <br/> `therefore` Molecular of the solution `=("No. of moles of sugar")/("Weight of water in kg")` <br/> `=0.1/0.18 = 0.56 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> kg^(-1)` <br/> Hence, the <a href="https://interviewquestions.tuteehub.com/tag/molal-1100217" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAL">MOLAL</a> concentration of the given solution is 0.56 m. <br/> (ii) Calculation of mole fraction : <br/> Number of moles of sugar `(n_(1))=0.1` <br/> Number of moles of water `(n_(2)) = 180/18 = 10` <br/> Total number of moles in solution `=n_(1) + n_(2) = 0.1 + 10 = 10.1` <br/> `therefore` The mole fraction of sugar `=n_(1)/(n_(1)+ n_(2)) = 0.1/10.1 = 9.9 xx 10^(-3)`</body></html>


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