Asucrose solution is prepared to a final concentration of 0.210 m . convert this value into terms of g/l, molality, and mass % (molecular weight, mwsucrose = 342.296 g/mol ; density, ρsol′n = 1.02 g/ml ; mass of water, mwat = 948.1 g ). note that the mass of solute is included in the density of the solution.

0.210 M sucrose solution is equivalent to 0.210 mol/ L sucrose solution. By multiplying it to the MW of sucrose, you will get 71. 8822 g/L sucrose solution. To get the molality, assume 1 L of solution. Multiplying 1 L to the density of the solution gives 1020 g of solution. Since there are 0.210 mol of sucrose, multiplying this to the MW of sucrose will give 71. 8822 g of sucrose. The mass of solvent can be obtained by subtracting the grams of solution from the grams of solute which is equal to 948.1178 g solvent. The molality is then equal to the mol of sucrose divided by kg of solvent which is 0.2215 molal. The mass percent is equal to the mass of solute divided by the mass of solution. The mass percent is 71.8822 g sucrose divided by 1020 g solution times 100, which is equal to 7.0473 mass %.

If it releases 5 tons a day and there are 7 days a take that answer and multiply by 5 percent (0.05) to find the amount released in one week. leave a comment if you need more .