What is the ratio by weight of NaF and NaI which when dissolved in water produces the same obtained on evaporation of the salt solution is 0.48 g per 100 mL of solution evaporataed. Assume complete dissociation of the salts.

What is the ratio by weight of NaF and NaI which when dissolved in wat

1.	What is the ratio by weight of NaF and NaI which when dissolved in water produces the same obtained on evaporation of the salt solution is 0.48 g per 100 mL of solution evaporataed. Assume complete dissociation of the salts.
Answer» Solution :Suppose in 100 mL of the solution, NaF = a G and NaI = b a . Then `a+b=0.48"…(i)"` `piV=i n R` For `NaF+NaI` solution in WATER, `"V = 100 mL = 0.1 L, n"=(a)/(42)+(b)/(150)` `""`(MOL. mass of NaF = `23+19=42,"Mol. mass of NaI"=23+127=150)` i = 2 (as NaF or NaI on dissociation produce 2 ions) `therefore""pixx0.1=2xx((a)/(42)+(b)/(150))xx0.0821xxT"...(II)"` For ureain water, `pi=CRT"(urea is non - electrolyte)"` `pi=0.1xx0.0821xxT"..(III)"` Dividing eqn. (ii) by eqn. (iii) `0.1=(2xx((a)/(42)+(b)/(150)))/(0.1)"or"(a)/(42)+(b)/(150)=(0.01)/(2)=0.005"or"150a+42b=31.5"...(iv)"` From eqns. (i) and (iv), on solving `a=0.105, b=0.375` `therefore"Ratio "=(a)/(b)=(0.105)/(0.375)=0.28`