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