if molality of glucosee in 1kg water is 1m then %age expression of solution is????

To determine the percentage concentration of a glucose solution when the molality is 1 m, we first need to understand the relationship between molality and percentage concentration. Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, we have 1 mole of glucose dissolved in 1 kg of water.

Understanding Molality

Molality is calculated using the formula:

• Molality (m) = moles of solute / kilograms of solvent

Since we have a molality of 1 m, it indicates that there is 1 mole of glucose in 1 kg of water. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol. Therefore, 1 mole of glucose weighs about 180 grams.

Calculating the Total Mass of the Solution

Next, we need to find the total mass of the solution, which includes both the solute (glucose) and the solvent (water).

• Mass of glucose = 180 g
• Mass of water = 1000 g (1 kg)

Thus, the total mass of the solution is:

• Total mass = Mass of glucose + Mass of water = 180 g + 1000 g = 1180 g

Finding the Percentage Concentration

Now, we can calculate the percentage by mass (weight percent) of the glucose in the solution using the formula:

• Percentage by mass = (mass of solute / total mass of solution) × 100

Substituting the values we have:

• Percentage by mass = (180 g / 1180 g) × 100

Calculating this gives:

• Percentage by mass ≈ 15.25%

Final Result

Therefore, the percentage concentration of the glucose solution, when the molality is 1 m, is approximately 15.25%. This means that in this solution, about 15.25% of the total mass is glucose, while the rest is water.