How many grams of solute are required to prepare 500ml of 0.5M urea solution,CO(NH2)2?
How many mols do you need?
mols = M x L = ?
Then mols = g/molar mass. You know molar mass and mols, solve for grams.
To calculate the number of grams of solute required to prepare a solution, we need to use the formula:
Mass of solute = molar concentration × volume of solution × molar mass of solute
First, let's determine the molar mass of urea (CO(NH2)2). Carbon (C) has a molar mass of 12.01 grams per mole (g/mol), nitrogen (N) has a molar mass of 14.01 g/mol, hydrogen (H) has a molar mass of 1.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Calculating the molar mass of urea:
(12.01 + 2 × 14.01 + 4 × 1.01) + 2 × (1.01 + 16.00)
= 60.06 g/mol
Now, we can calculate the number of grams of solute required. We have a molar concentration of 0.5 moles per liter (M = mol/L) and a volume of 500 mL, which we need to convert to liters.
1 L = 1000 mL, so 500 mL = 500/1000 = 0.5 L.
Mass of solute = 0.5 M × 0.5 L × 60.06 g/mol
= 0.25 × 60.06 g
= 15.015 g (rounded to three decimal places)
Therefore, to prepare a 500 mL solution with a concentration of 0.5 M of urea (CO(NH2)2), you would need approximately 15.015 grams of urea.