Question

Calculate the molarity of each of the following solutions:

(a) 0.195 g of cholesterol, C27H46O, in 0.100 L of serum, the average concentration of cholesterol in human serum

(b) 4.25 g of NH3 in 0.500 L of solution, the concentration of NH3 in household ammonia

(c) 1.49 kg of isopropyl alcohol, C3H7OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol

(d) 0.029 g of I2 in 0.100 L of solution, the solubility of I2 in water at 20 °C

Please show me above answers step by step...

thanks...

Answer 1

a) M = (w/W)*(1/V in L)

   w = wt of solute = 0.195 g

   W = molarmass of cholesterol = 386.65 g/mol

   V = Vol of solution = 0.1 L

       = (0.195/386.65)*(1/0.1)

       = 0.005 M

b) M = (w/W)*(1/V in L)

   w = wt of solute = 4.25 g

   W = molarmass of NH3 = 17 g/mol

   V = Vol of solution = 0.5 L

       = (4.25/17)*(1/0.5)

concentration of NH3 = 0.5 M

c) M = (w/W)*(1/V in L)

   w = wt of solute = 1.49 kg = 1490 g

   W = molarmass of C3H7OH = 60 g/mol

   V = Vol of solution = 2.5 L

       = (1490/60)*(1/2.5)

   concentration of rubbing alchol = 9.93 M

D) M = (w/W)*(1/V in L)

   w = wt of solute = 0.029 g

   W = molarmass of I2 = 253.809 g/mol

   V = Vol of solution = 0.1 L

       = (0.029/253.809)*(1/0.1)

concentration of I2 = 0.00114 M