Which relations are correct for an aqueous dilute solution of $K_{3} {PO}_{4}$ if its degree of dissociation $α$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${Mol.wt.of K}_{3} {PO}_{4} = {Mol.wt.}_{obs.} \times (1 + 3 α)$

$\frac{Δ P}{P^{°}} = \frac{molality \times 18 \times (1 + 3 α)}{1000}$

$\frac{Δ P}{P^{°}} = \frac{π_{obs.} \times 18 \times (1 + 3 α)}{ST \times 1000}$

$\frac{Δ P}{P^{°}} = \frac{Δ T_{fobs.} \times 18}{K_{f} \times 1000}$

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{Δ P}{P^{0}} = \frac{n}{N} = \frac{n \times M \times 1000}{W \times 1000} = \frac{molality \times M}{1000} For electrolyte \frac{Δ P}{P^{0}} = \frac{molality \times M}{1000} \times (1 + 3 α) (M = 18 for H_{2} O) Also, π_{obse.} = C \times S \times T (1 + 3 α) Δ T_{f obse.} = K_{f} \times m o l a l i t y \times (1 + 3 α) i = (1 + 3 α) = \frac{C a l . M . w t}{Obser . M . w t} ∴ {M.Wt.of K}_{3} {PO}_{4} = M_{obsc.} \times (1 + 3 α)$