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

Answer

(i) 0.556 (ii) 0.0099

Explanation

(i) 0.556 (ii) 0.0099
214.2 gm solution has 34.2 gm sugar (sugar is sucrose,`C_(12)H_(22)O_(11)` )
180 gm `H_(2)O` has `(34.2)/(342) = 0.1` mole sugar
1000 gm `H_(2)O` would have `= 0.1 xx (1000)/(180)` mole sugar `rArr 0.556` molal
`chi_("sugar") = (n_("sugar"))/(n_("sugar") + n_(H_(2)O)) = (0.1)/(0.1+(180)/(18)) = (0.1)/(10.1) = 0.0099`