200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be2.52xx 10^(-3)bar.The molar mass of protein will be (R=0.083 Lbar mol ^(-1) K^(-1))

200 ml of an aqueous solution of a protein contains 1.26 g of protein.

1.	200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be2.52xx 10^(-3)bar.The molar mass of protein will be (R=0.083 Lbar mol ^(-1) K^(-1))
Answer» <html><body><p>`62.22` <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> `mol^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`<br/>`12444 g mol ^(-1)`<br/>`300 g mol ^(-1)`<br/>none of these </p>Solution :`pi = CRT` <br/> `pi = (<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)/(M xxV) xx RT` <br/> ` therefore M = (<a href="https://interviewquestions.tuteehub.com/tag/wrt-1462282" style="font-weight:bold;" target="_blank" title="Click to know more about WRT">WRT</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/piv-591561" style="font-weight:bold;" target="_blank" title="Click to know more about PIV">PIV</a>) = (1.26 xx 0.083 xx 300)/(2.52 xx 10 ^(-3) xx 0.2)= 62.22 kg mol ^(-1)`</body></html>