Nitric acid is commonly purchased as a 16 M solution, which has a density of 1.41 g/mL.
Convert this molarity to a) mass percentage, b) molality, c) mole fraction.
A) 16molHNO3(63.02HNO3/1MolHNO3)=1,008.32gHNO3
- 1,000gH20(1MolH20/1.41ml/g)=1410ml/mol
w/w= 1008.32/1410= 71.5%w/w
B)1410-1008.32= 401.88 --> 16molHNO3/401.32= 0.40m
C)71.5%w/w (1mol/63.02)= 1.13--> 1.13
100gSoln(1ml/1.41)= 70.92--> .113/.71= .16 mol frac
A and B are ok.
I think C is incorrect.
If the soln is 71.5%, then you are right that 1.13 mol HNO3.
100 g soln = 71.5g HNO3 = 2.85g H2O (I think you converted to volume).
Then XHNO3 = 1.13 mols HNO3/(1.13+2.85) = about 0.4 or so.
Dr.Bob,
How did you obtain 2.85gH20?
I made two typos and did not show a step. Here it is in detail.
That should be 100g total - 71.5g HNO3 = 28.5 H2O.
mols H2O = 28.5/18 = 1.58
total mols = 1.58 + 1.13 = 2.71
XHNO3 = 1.13/2.71 = 0.417
To convert the molarity of nitric acid to mass percentage, you need to find the mass of the solute (HNO3) and the mass of the solution (HNO3 + H2O).
a) First, calculate the mass of HNO3:
16 mol of HNO3 x (63.02 g/mol of HNO3) = 1,008.32 g of HNO3
Next, calculate the mass of the solution:
1,000 g of H2O x (1 mol of H2O / 1.41 mL of H2O) = 1410 mL of H2O per mol
To find the mass percentage, divide the mass of HNO3 by the mass of the solution and multiply by 100:
Mass percentage = (1,008.32 g of HNO3 / 1410 g of solution) x 100 = 71.5% w/w
b) To convert the molarity to molality, subtract the mass of HNO3 from the mass of the solution:
1410 g of solution - 1008.32 g of HNO3 = 401.88 g of H2O
Then, divide the moles of HNO3 by the mass of H2O in kg:
16 mol of HNO3 / 401.88 g of H2O = 0.04 molality (mol/kg)
c) To calculate the mole fraction, divide the moles of HNO3 by the total moles of solute and solvent:
Mole fraction = (0.04 mol of HNO3 / 63.02 g/mol of HNO3) / ((100 g of solution / 1.41) + (0.04 mol of HNO3 / 63.02 g/mol of HNO3)) = 0.16 mole fraction