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  • 2.5 g of protein containing only glycine <img alt="left(mathrm{C}_{2} mathrm{H}_{5} mathrm{NO}_{2} ight)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28%5Cmathrm%7BC%7D_%7B2%7

2.5 g of protein containing only glycine left(mathrm{C}_{2} mathrm{H}_{5} mathrm{NO}_{2} ight) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 imes 10^{-3} bar. The total number of glycine units present in the protein is__________.

ext { (Given : } mathrm{R}=0.083 mathrm{~L} ext { bar } mathrm{K}^{-1} mathrm{~mol}^{-1} ext { ) } 

Option: 1

330


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given, \text{T=300K}
           \begin{aligned} &\mathrm{P=5.03 \times 10^{-3} \; \mathrm{bar} }\\ &\mathrm{R=0.083 \mathrm{~L}\; \mathrm{bar\: K}^{-1} \mathrm{mol}^{-1}} \end{aligned}

Osmotic pressure -


        \begin{gathered} \mathrm{\pi=C R T} \\\\ 5.03 \times 10^{-3}=\text{C}(0.083)(300) \\\\ \text{C}=0.202 \times 10^{-3} \mathrm{M} \end{gathered}


\mathrm{moles\: of \: protein =0.202 \times 10^{-3} \times 0.5 }

                                    =10^{-4}\times 1.01

We have-

            \mathrm{\frac{\text { given mass }}{\text { molar mass }}=n o .of\: moles}

    \therefore \frac{2.5}{\text{M}}=1.01\times 10^{-4}

\mathrm{M(molar\: mass\: of\: protein)}=24752

\text{no of glycerine units}=\frac{24752}{75}=330.03

Hence, 330 is correct answer

Posted by

Gaurav

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