How much by mass of Urea (NH2)2CO is required for a 32.5% solution in one litre of water.
A 32.5% w/v (is that what you want) soln is 32.5g/100 mL; therefore, it will be 325 g/L.
To find out how much urea (NH2)2CO is required to make a 32.5% solution in one liter of water, we first need to understand what a 32.5% solution means.
A 32.5% solution means that in one liter of the resulting solution, there will be 32.5 grams of urea and the remaining mass will be water.
To calculate the mass of urea required, we will use the formula:
Mass of Urea = Total solution volume (in liters) × Percentage concentration × Density
Let's assume the density of the solution is 1 g/mL, as the density of water is close to 1 g/mL.
Given:
Total solution volume = 1 liter
Percentage concentration = 32.5%
Using the formula:
Mass of Urea = 1 liter × 0.325 × 1 g/mL = 0.325 g
Therefore, you would need 0.325 grams of urea ((NH2)2CO) to make a 32.5% solution in one liter of water.