JEE Advance - Chemistry (1988 - No. 5)
A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12H22O11). Calculate (i) molal concentration and (ii) mole fraction of sugar in the syrup
(i) 0.56, (ii) 0.0099
(i) 0.65, (ii) 0.0199
(i) 0.56, (ii) 0.1099
(i) 1.56, (ii) 0.0099
(i) 0.50, (ii) 0.0050
Selitys
Weight of sugar syrup = 214.2 gm
Weight of solute sugar = 34.2 gm
$$\therefore$$ Weight of solvent (H2O) = 214.2 $$-$$ 34.2 = 180 gm
(i) Molal concentration or molality
$$ = {{Number\,of\,moles\,of\,solute} \over {Weight\,of\,solvent\,in\,kg}}$$
$$ = {{{{34.2} \over {342}}} \over {{{180} \over {1000}}}}$$
$$ = {{100} \over {180}} = 0.56$$
(ii) Mole fraction of sugar
$$ = {{{{34.2} \over {342}}} \over {{{34.2} \over {342}} + {{180} \over {18}}}}$$
$$ = {{0.1} \over {0.1 + 10}}$$
$$ = {{0.1} \over {10.1}} = 0.0099$$