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0.200g of iodine is stirred in 100mL of water. After equilibrium is reached, we add 150mL of water to the system. How much iodine will be left undissolved?

Answer» <html><body><p>`1.3g`<br/> `0.130g`<br/> `0.013g`<br/> `13g`</p>Solution :After an equilibrium between `0.200g` of iodine and `100mL` of water is reached, the <a href="https://interviewquestions.tuteehub.com/tag/addition-367641" style="font-weight:bold;" target="_blank" title="Click to know more about ADDITION">ADDITION</a> of `<a href="https://interviewquestions.tuteehub.com/tag/150ml-1791979" style="font-weight:bold;" target="_blank" title="Click to know more about 150ML">150ML</a>` of water will result in the further dissolution of `I_2`. <br/> <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of solution after the addition of `150mL` of water <br/> `=100mL+150mL` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/250ml-297852" style="font-weight:bold;" target="_blank" title="Click to know more about 250ML">250ML</a>` <br/> According, `100mL` of solution contains `0.28g I_2`. <br/> `1mL` solution contains `(0.28g)/(<a href="https://interviewquestions.tuteehub.com/tag/1000ml-1771304" style="font-weight:bold;" target="_blank" title="Click to know more about 1000ML">1000ML</a>)` <br/> Therefore, `250mL` of solution will contain <br/> `(0.28g)/(1000ml)xx250mL=0.070g` of `I_2` <br/> Undissolved iodine `=(0.200g)-(0.070g)` <br/> `=0.130g`</body></html>


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