Answer:
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
