Question

23 g of C12H22O11 (sugar) was dissolved in 54 gm of H2O. Calculate :

(a) mole fraction of sugar. (2 M)

(b) mole fraction of H2O (1 M)

(c) molality (2 M)​

Answer 1

Answer:

no. of moles of sugar =23÷342=0.06 moles

no.of moles of water=54÷18=3moles

a)mole fraction of sugar=0.06÷3+0.06==0.019

b)mole fraction of water=3÷3+0.06=0.98

Answer 2

Given :

• Mass of C12H22O11 = 23g
• Mass of H2O = 54g

To find :

(a) Mole fraction of sugar

(b) Mole fraction of H2O

(c) Molality

Formula used :

• mole fraction of a = number of mole of a ÷ ( total number of mole in solution)

• molality = number of mole of solute ÷ mass of solvent (in Kg)

Important point :

• molar mass of sugar (C12H22O11) = 342g/mol
• molar mass of H2O = 18g/ml

Solution :

• moles of sugar = mass of sugar ÷ molar mass of sugar

mole of sugar = 23 ÷ 342 = 0.06

• moles of H2O = mass of H2O ÷ molar mass of H2O

moles of H2O = 54 ÷ 18 = 3

• mass of H2O (in Kg) = 54 ÷ 1000 = 0.054kg

(a)

mole fraction of sugar = moles of sugar ÷ ( mole of sugar + moles of H2O)

mole fraction of sugar = 0.06 ÷ (0.06+3)

mole fraction of sugar = 0.019

(b)

mole fraction of H2O = moles of H2O ÷ ( moles of sugar + moles of H2O )

mole fraction of H2O = 3 ÷ (3+0.06)

mole fraction of H2O = 0.981

(c)

molarity = number of mole of sugar ÷ mass of H2O (in Kg)

molality = 0.06 ÷ 0.054

molality = 1.11m

ANSWER :

(a) 0.019

(b) 0.981

(c) 1.11 m