Solving for 16M HN03 with a Density of 1.42g/ml

[hockeynicole]

Homework Statement

16 M
1.42g/ml density
formula HN03

Homework Equations

The Attempt at a Solution

 

[symbolipoint]

16 moles/1000 ml. You find the formula weight of HNO3 yourself; call the result, FW for "formula weight";
(16 moles *FW grams/mole)/(1000 ml.)
Can you do the rest?

 

[Blueshift5]

To solve for the mass of 16M HN03, we can use the formula: M = D x V, where M is the mass, D is the density, and V is the volume. In this case, we are given the density of 1.42g/ml, so we can rewrite the equation as M = 1.42g/ml x V. However, we still need to determine the volume in order to find the mass.

To find the volume, we can use the formula: M = n x Mw, where M is the molarity, n is the number of moles, and Mw is the molar mass. Since we know the molarity (16M) and the molar mass of HN03 (63.01 g/mol), we can rearrange the equation to solve for n: n = M/Mw. Plugging in our values, we get n = 16M/63.01 g/mol = 0.254 mol.

Now, we can use the ideal gas law, PV = nRT, to calculate the volume. However, since we are dealing with a liquid and not a gas, we can use the formula for the volume of a liquid: V = n/Mw x (RT/P). Plugging in our values, we get V = 0.254 mol/63.01 g/mol x (0.08206 L.atm/mol.K x 298 K)/1 atm = 0.0103 L.

Finally, we can plug in our calculated volume into the first equation to solve for the mass: M = 1.42g/ml x 0.0103 L = 0.0142 g. Therefore, the mass of 16M HN03 with a density of 1.42g/ml is 0.0142 g.