A sample of conc. nitric acid has a density of 1.41g/ml and contains 70% HNO3 by mass.
What mass of HNO3 is present per L of solution?
What is the molarity?
Assume 1 liter. then mass of acid is .7*1.41*1000, and from that calculate moles acid
Volume is 1liter,
molarity=molesacid/1
s12
.987 g HNO3
M=.0157
To find the mass of HNO3 per liter of solution, we first need to calculate the mass of the solution in one liter.
Since the density of the solution is given as 1.41 g/ml, we know that one milliliter of the solution has a mass of 1.41 grams. Therefore, one liter (1000 ml) of the solution has a mass of:
Mass of solution = density * volume = 1.41 g/ml * 1000 ml = 1410 grams
Now we can determine the mass of HNO3 present in one liter of the solution. The concentration of HNO3 is given as 70% by mass. This means that 70 grams of HNO3 are present in 100 grams of the solution.
Mass of HNO3 in one liter of solution = 70% * 1410 grams = 0.70 * 1410 grams = 987 grams
Finally, to find the mass of HNO3 per liter of solution, we divide the mass of HNO3 by the volume of the solution:
Mass of HNO3 per liter of solution = 987 grams / 1 liter = 987 grams/L
To calculate the molarity of the solution, we need to convert the mass of HNO3 to moles. The molar mass of HNO3 is 1 + 14 + 48 = 63 g/mol.
Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3 = 987 grams / 63 g/mol = 15.67 moles
Finally, we divide the moles of solute by the volume of the solution in liters to find the molarity:
Molarity = Moles of solute / Volume of solution in liters = 15.67 moles / 1 liter = 15.67 M
Therefore, the mass of HNO3 per liter of solution is 987 grams/L, and the molarity of the solution is 15.67 M.