2.5 g of protein containing only glycine ( C 2 H 5 NO 2) 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 × 10-3 bar. The total number of glycine units present in the protein is (Given: R =0.083 L bar K -1 mol -1 )

Question

2.5 g of protein containing only glycine ( C 2 H 5 NO 2) 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 × 10-3 bar. The total number of glycine units present in the protein is (Given: R =0.083 L bar K -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

π= CRT 5.03 × 10-3= C × 0.083 × 300 C =0.202 × 10-3 M Moles of protein =0.202 × 10-3 × 0.5 =10-4 × 1.01 1.01 × 10-4 × (2.5/ M ) M(molar mass of protein )=24752 ∴ No. of glycine units =(24752/75)=330.03

Q. 2.5g of protein containing only glycine (C2​H5​NO2​) is dissolved in water to make 500mL of solution. The osmotic pressure of this solution at 300K is found to be 5.03×10−3 bar. The total number of glycine units present in the protein is ___ (Given : R=0.083L bar K−1mol−1 )

π=CRT 5.03×10−3=C×0.083×300 C=0.202×10−3M Moles of protein =0.202×10−3×0.5 =10−4×1.01 1.01×10−4×M2.5​ M(molar mass of protein )=24752 ∴ No. of glycine units =7524752​=330.03