Solution.
K=C(I2 in H2O)C(I2 in CCl4)K = \frac{C(I2 \ in \ H2O)}{C(I2 \ in \ CCl4)}K=C(I2 in CCl4)C(I2 in H2O)
C(I2 in org ph)=17∗C15=1.133CC(I2 \ in \ org \ ph) = \frac{17*C}{15} = 1.133 CC(I2 in org ph)=1517∗C=1.133C
C(I2 in water ph)=22∗C∗0.250=0.088CC(I2 \ in \ water \ ph) = \frac{22*C*0.2}{50} = 0.088 CC(I2 in water ph)=5022∗C∗0.2=0.088C
C1=1.133C∗5015=3.78CC1 = \frac{1.133C*50}{15} = 3.78 CC1=151.133C∗50=3.78C
C2=0.088C∗15050=0.264CC2 = \frac{0.088C*150}{50} = 0.264 CC2=500.088C∗150=0.264C
K=0.264C3.78C=0.070K = \frac{0.264C}{3.78C} = 0.070K=3.78C0.264C=0.070
Answer:
K = 0.070
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Dear Claudiu, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.
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