The mass of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of urea in the same quantity of water is (A) 1 g (B) 3 g (C) 6 g (D) 18 g

To determine the mass of glucose needed to achieve the same lowering of vapor pressure as 1 g of urea dissolved in 50 g of water, we can use Raoult's Law. This law states that the vapor pressure of a solvent decreases when a solute is added, and the extent of this decrease is proportional to the number of solute particles in the solution. Let's break down the problem step by step.

Understanding the Components

First, we need to recognize that both urea and glucose are solutes that will dissociate differently in solution. Urea (NH₂CONH₂) is a non-electrolyte and does not dissociate into ions, while glucose (C₆H₁₂O₆) also does not dissociate. Therefore, the number of moles of each solute will directly affect the lowering of vapor pressure.

Calculating Moles of Urea

To find out how many moles of urea are in 1 g, we use its molar mass. The molar mass of urea is approximately 60 g/mol. Thus, the number of moles of urea can be calculated as follows:

• Moles of urea = mass (g) / molar mass (g/mol)
• Moles of urea = 1 g / 60 g/mol = 0.01667 moles

Relating Moles of Glucose

Next, we need to find out how many moles of glucose would produce the same lowering of vapor pressure. Since both solutes do not dissociate, we can directly compare the moles. Therefore, we need 0.01667 moles of glucose as well.

Calculating Mass of Glucose

The molar mass of glucose is approximately 180 g/mol. Now, we can find the mass of glucose required to have 0.01667 moles:

• Mass of glucose = moles × molar mass
• Mass of glucose = 0.01667 moles × 180 g/mol = 3 g

Final Answer

Thus, to achieve the same lowering of vapor pressure as dissolving 1 g of urea in 50 g of water, you would need to dissolve 3 g of glucose. Therefore, the correct answer is (B) 3 g.