Commercial nitric acid comes in a concentration of 16.0 mol/l. the density of this solution is 1.42 g/ml
calculate the percent (w/w) of nitric acid, hno3, in this solution.
how many milliliters of the concentrated acid have to be taken to prepared 250 g of a solution that is 10.0%(w/w) hno3?

71.0% (w/w)

24.8 mL

Explanation:

First, let's calculated the molar mass of the acid, knowing the molar masses of its elements:

H = 1 g/mol

N = 14 g/mol

O = 16 g/mol

So, the molar mass is: 1x1 + 1x14 + 3x16 = 63.0 g/mol

In 1 L of the solution, there is 16.0 mol of the acid, so the mass will be the number of moles multiplied by the molar mass:

m = 16.0x63.0 = 1008.0 g

The density is 1.42 g/mol, so the total mass of the solution is the density multiplied by the volume. For 1 L (1000 mL):

mt = 1.42x1000 = 1420.0 g

The percent of nitric acid is its mass divided by the total mass multiplied by 100%

(1008.0/1420.0)x100% = 71.0% (w/w)

To prepare a solution by dilution of the other solution, we can use the equation:

C1m1 = C2m2

Where C is the concentration, m is the mass, 1 is the initial solution and 2 the final solution. So:

71%xm1 = 10%x250

m1 = 35.21 g

The volume is the mass divided by the density:

V = 35.21/1.42

V = 24.8 mL

i would say b

explanation:

but i think i would need more information than that