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.
Text Solution
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The correct Answer is:
`(9.9 xx 10^(-3))`
Moles of sugar `=34.2/342=0.1` Moles of water in syrup`=214.2-34.2` `=180 g` Therefore, (i) Molarity `=("Moles of solute")/("Weight of Solvent (g)")xx1000` `=0.1/180xx1000=0.55` (ii) Mole fraction of sugar `=("Mole of sugar")/("Mole of sugar + Mole of water")` `=0.1/(0.1+10)=9.9xx10^(-3)`
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