Question

1. What volume of 2.25 M nitric acid is needed to convert 9.0 g of m-chlorophenol to
3-chloro-4,6-dinitrophenol? (No excess of nitric acid is used.)

Answer 1

3-chloro-4,6-dinitrophenol is C6H3ClN2O5

Molar mass of C6H3ClN2O5,

MM = 6*MM(C) + 3*MM(H) + 1*MM(Cl) + 2*MM(N) + 5*MM(O)

= 6*12.01 + 3*1.008 + 1*35.45 + 2*14.01 + 5*16.0

= 218.554 g/mol

mass(C6H3ClN2O5)= 9.0 g

number of mol of C6H3ClN2O5,

n = mass of C6H3ClN2O5/molar mass of C6H3ClN2O5

=(9.0 g)/(218.554 g/mol)

= 4.118*10^-2 mol

for 1 mol of C6H3ClN2O5, 2 mol of HNO3 is required

So,

mol of HNO3 required = 2*4.118*10^-2 mol

= 8.236*10^-2 mol

Now use:

mol of HNO3 = M(HNO3)*V(HNO3)

8.236*10^-2 mol = 2.25 M * V

V = 0.0366 L

V = 36.6 mL

Answer: 36.6 mL