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[Monk G]

An aqueous solution of glucose, C_2 H_12 O_6 has a mole fraction of 0.4 for glucose. What is the molality of this solution?
A. 31m B. 46m
C. 37m D. 54m

 

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To find the molality of the solution, we need to know the number of moles of solute (glucose) and the mass of the solvent (water).

Given that the mole fraction of glucose (C₂H₁₂O₆) is 0.4, it means that 40% of the total moles in the solution are glucose.

Let's assume we have 100 moles of solution. Since the mole fraction of glucose is 0.4, the moles of glucose in the solution would be 0.4 * 100 = 40 moles.

Now, we need to find the mass of water in the solution. To do that, we need to subtract the mass of glucose from the total mass of the solution.

The molar mass of glucose (C₂H₁₂O₆) can be calculated as follows:
(2 * Atomic mass of carbon) + (12 * Atomic mass of hydrogen) + (6 * Atomic mass of oxygen)
= (2 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol)
= 24.02 g/mol + 12.12 g/mol + 96.00 g/mol
= 180.14 g/mol

Since we have 40 moles of glucose, the mass of glucose in the solution would be 40 moles * 180.14 g/mol = 7205.6 g.

The total mass of the solution is not given, so let's assume it is 1000 g for simplicity.

Therefore, the mass of water in the solution would be 1000 g - 7205.6 g = -6205.6 g. This negative value is not possible, indicating an error in the information given. Please double-check the values provided and re-calculate accordingly.