Calculate the amount of urea required to prepare 250 mL of a 1.55 M solution. The molecular weight (MW) of urea is 60.06 g/mol. Represent the amount in both grams (g) and milligrams (mg).

To prepare a 250 ml (0.250 L) solution of 1.55 M urea, 23.27335 grams or 23273.35 milligrams of urea, which has a molar mass of 60.06 g/mol, is needed.

Explanation:

To calculate the amount of urea (CO(NH2)2) required to prepare 250 ml of a 1.55 m solution, we first need to find the number of moles of urea required using the formula:

Molarity (M) = moles of solute / liters of solution

Since we have the molarity (1.55 m) and the volume of the solution (0.250 L), we can rearrange this formula to solve for moles of urea:

moles of urea = Molarity \$ ext{M}\$ \u00d7 Volume of solution \$ ext{L}\$

moles of urea = 1.55 moles/L \u00d7 0.250 L = 0.3875 moles

To find the mass of urea required, we then use the mw of urea = 60.06 g/mol. The mass can be calculated by multiplying the number of moles by the molar mass:

mass of urea = moles \u00d7 molar mass

mass of urea = 0.3875 moles \u00d7 60.06 g/mol = 23.27335 g

To convert to milligrams (mg), we know that 1 g = 1000 mg, so it follows that:

23.27335 g \u00d7 1000 mg/g = 23273.35 mg

We can now conclude that the amount of urea required is 23.27335 g or 23273.35 mg.