KCET2014ChemistryAldehydes and Ketones
( 1.78 ~g ) of an optically active ( L )-amino acid ( ( A ) ) is treated with ( N a ~N O ₂ / H Cl ) at ( 0^ C .448 ~cm ³ ) of nitrogen was at STP is evolved. A sample of protein has ( 0.25 % ) of this amino acid by mass. The molar mass of the protein is
Options
- A( 34,500 ~g ~mol ⁻¹ )
- B( 35,600 ~g ~mol ⁻¹ )
- C( 36,500 ~g ~mol ⁻¹ )
- D( 35,400 ~g ~mol ⁻¹ )
Correct answer
B. ( 35,600 ~g ~mol ⁻¹ )
Step-by-step solution
Mass of L -amino acid =178 ~g R- CH - COOH - Na ₂ / HCl NH ₂ At STP , 1 ~mol =22400 ~cm ³ C 1 ~cm ³= 1 22400 ~mol So, 448 ~cm ³ ~N ₂ is evolved when 448 22400 ~mol of amino acid reacted Now, for L -amino acid, Molar mass of amino acid = 22400 1.78 10² 448 = 178 2 =89 gmol ⁻¹ Sample of protein contains 0.25 % amino acid, so 100 ~g of protein contain 0.25 ~g of amino acid. Therefore, Molecular mass of protein = 100 89 0.25 =35600 gmol ⁻¹