When 34.2 g of cane sugar C12H22O11 is dissolved in water the solution weighs 214.2 g What is the molality of the solution?

CBSE 12-science - Chemistry

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When 34.2 g of cane sugar C12H22O11 is dissolved in water the solution weighs 214.2 g What is the molality of the solution?

Asked by manishachand010 | 27 May, 2022, 10:14: AM

When 34.2 g of cane sugar C12H22O11 is dissolved in water the solution weighs 214.2 g What is the molality of the solution?
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When 34.2 g of cane sugar C12H22O11 is dissolved in water the solution weighs 214.2 g What is the molality of the solution?

Asked by rufinshafeek | 13 May, 2020, 02:54: PM

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When 34.2 g of cane sugar C12H22O11 is dissolved in water the solution weighs 214.2 g What is the molality of the solution?

Text Solution

Solution : Weight of sugar syrup = 214.2 g <br> Weight of sugar present = 34.2 g <br> Molecular mass of sugar `(C_(12)H_(22)O_(11)) = 342` <br> (i) Calculation of molality : <br> Number of moles of sugar = `34.2/342 = 0.1` <br> Weight of water `=214.2 - 34.2 = 180 g = 0.18 g` <br> `therefore` Molecular of the solution `=("No. of moles of sugar")/("Weight of water in kg")` <br> `=0.1/0.18 = 0.56 mol kg^(-1)` <br> Hence, the molal concentration of the given solution is 0.56 m. <br> (ii) Calculation of mole fraction : <br> Number of moles of sugar `(n_(1))=0.1` <br> Number of moles of water `(n_(2)) = 180/18 = 10` <br> Total number of moles in solution `=n_(1) + n_(2) = 0.1 + 10 = 10.1` <br> `therefore` The mole fraction of sugar `=n_(1)/(n_(1)+ n_(2)) = 0.1/10.1 = 9.9 xx 10^(-3)`

214.2 g of sugar syrup contains 34.2 g of sugar. Calculate (i) molarity of the solution (ii) mole fraction of sugar in the syrup. Chemistry Q&A

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