Question

Rajat prepared 100 g of aq. NaCl solutioncontaining NaCl 40% by mass. He divided thesolution into two equal parts, the mole fractionof NaCl in each part is [Molar mass of NaCl =58.5 g/mol)(1) 0.17(2) 0.5(3) 0.4(4) 0.34​

Answer 1

Answer:

100 × 40/100 = 40g of NaCl

40g of NaCl is in the solution

Rest 60g is water/solvent

Moles of NaCl = 40/58.5 = 0.68

Moles of H20 = 60/18 = 3.33

Total moles = 4.01

Moles of NaCl in each part = 0.68/2 = 0.34

Moles of H2O in each part = 3.33/2 = 1.67

Total moles in each part = 0.68 + 0.34 = 2.01

Mole fraction of NaCl in each part = 0.34/2.01 = 0.169 or 0.17

Thus the answer is (1) 0.17

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Answer 2

Answer : - 3274.2 kJ/mol

Explanation: moles present = 0.532/78 = 0.00682

dU = -22.4/ 0.00682 = - 32469.8 kJ/mole

As you know

dH = dU + dngRt

= -3269.8 + 1.5 × 8.314/1000 ×353

= -3274.2 kJ/mole