\(\Delta T_f = 0.186^o\) સે
હવે, \(\Delta T_f = K_f × m\)
\(0.186 = 1.86 \times m\)
\(m = 0.1\)
\(\Delta T_b = K_b × m = 0.512 \times 0.1 = 0.0512^o\) સે
$(i)$ $0.10\, {M} \,{Ba}_{3}\left({PO}_{4}\right)_{2}$
$(ii)$ $0.10\, {M}\, {Na}_{2} {SO}_{4}$
$(iii)$ $0.10\, {M}\, {KCl}$
$(iv)$ $0.10 \,{M} \,{Li}_{3} {PO}_{4}$