1.

What is the molality of a dilute aqueous 0.02 N H3PO4 solution?(a) 0.0050(b) 0.0200(c) 0.00330(d) 0.0067I got this question during an online exam.The above asked question is from Expressing Concentration of Solutions topic in chapter Solutions of Chemistry – Class 12

Answer»

The correct ANSWER is (d) 0.0067

The explanation: Given,

Normality, N = 0.02

It is known that N = nf x M, where nf is the n-factor of equivalence and M is the molarity of the solution.

On expanding, N = nf x \(\frac{Number \, of \, moles}{VOLUME \, of \, solution(L)}\)

For H3PO4, nf = 3 since there are 3 breakable OH bonds.

Therefore, N = 3 x \(\frac{Number \, of \, moles}{mass \, of \, solvent(kg)}\) . Here, volume of solution (1 L) can be approximated as mass of solvent (1 kg) since it is an dilute solution in water (density = 1 kg/L) implying water concentration is far greater such that amount of H3PO4 is negligible.

Now, \(\frac{Number \, of \, moles}{mass \, of \, solvent(kg)}\) = molality of solution

Thus, N/3 = 0.02/3 =0.0067 MOLAL.


Discussion

No Comment Found

Related InterviewSolutions