The amount of urea in 100g water of 0.3 molal solution is

Explanation

We are given a 0.3 molal solution of urea in 100g of water. Molality (m) is defined as the number of moles of solute (urea) per kilogram of solvent (water). We need to find the amount of urea (solute) in grams present in 100g of water (solvent).Here's how to calculate the mass of urea:Convert grams of solvent to kilograms:We have 100g of water.To convert grams to kilograms, we divide by 1000:100g / 1000 g/kg = 0.1 kgCalculate the moles of urea:We are given a molality of 0.3 m.Molality (m) = moles of solute (urea) / mass of solvent (kg)Therefore, moles of urea = molality (m) * mass of solvent (kg)moles of urea = 0.3 mol/kg * 0.1 kg = 0.03 molConvert moles of urea to grams:We know the molar mass of urea is 60.06 g/mol (you can find this information in the periodic table or a reference book).Mass of urea = moles of urea * molar massMass of urea = 0.03 mol * 60.06 g/mol = 1.802 g (round to two decimal places)Therefore, there are approximately 1.8 grams of urea in 100g of water in the 0.3 molal solution.