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How many m. moles of sourose should be dissolved in 500 grams of water so as to get a solution which has a difference of `103.57^@C` between boiling point and freezing point ? `(K_f=1.86 K kg "mol"^(-1), K_b=0.52 K Kg "mol"^(-1))`A. 500 m. molesB. 900 m. molesC. 750 m. molesD. 650 m. moles |
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Answer» Correct Answer - C Boiling point of solution =boiling point `+DeltaT_b=100+DeltaT_b` Freezing point of solution =freezing point `-DeltaT_f=0-DeltaT_f` Difference in temperature (given)=`100+DeltaT_b-(-DeltaT_f)` `103.57=100+DeltaT_b+DeltaT_b+DeltaT_f=100+"molality"xxK_b+"molality"xxK_f` `=100+"molality"(0.52+1.86)` Molality=`(103.57-100)/238=3.57/2.38=1.5 m` and molality =`("moles"xx1000)/W_(gm(solvent)), 1.5=("moles"xx1000)/500` Moles of solute =`(1.5xx500)/1000=0.75` moles |
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