26.65gm of compound having formula ([Cr (H2 O)5 Cl)](Cl)2⋅ H2O is mixed with 498.2gm water. If complex is 75 % ionised in water, then elevation in boiling point in Kelvin for above solution is X . Calculate 4X ? ( k b . water =0.50 k kg mol -1 ) (At. wt. .Cr: 52, Cl: 35.5, O: 16)

Question

26.65gm of compound having formula ([Cr (H2 O)5 Cl)](Cl)2⋅ H2O is mixed with 498.2gm water. If complex is 75 % ionised in water, then elevation in boiling point in Kelvin for above solution is X . Calculate 4X ? ( k b . water =0.50 k kg mol -1 ) (At. wt. .Cr: 52, Cl: 35.5, O: 16) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(26.65/266.5)=0.1 mole Total water =498.2+18=500 gm beginarrayl Δ T b = imk b =(1+2 × (3/4)) (0.1/0.5) × 0.5 =0.254 X =4 × 0.25=1 endarray

Q. 26.65gm of compound having formula ([Cr(H2​O)5​Cl)](Cl)2​⋅H2​O is mixed with 498.2gm water. If complex is 75% ionised in water, then elevation in boiling point in Kelvin for above solution is X . Calculate 4X ? (kb​ water =0.50kkgmol−1 ) (At. wt. Cr:52,Cl:35.5,O:16)

266.526.65​=0.1 mole Total water =498.2+18=500gm ΔTb​=imkb​=(1+2×43​)0.50.1​×0.5=0.254X=4×0.25=1​

$\frac{26.65}{266.5} = 0.1$ mole
Total water $= 498.2 + 18 = 500 g m$
$Δ T_{b} = im k_{b} = (1 + 2 \times \frac{3}{4}) \frac{0.1}{0.5} \times 0.5 = 0.254 X = 4 \times 0.25 = 1$