1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 × 10-3 bar The molar mass of the biopolymer is -104 gmol -1 (Round off to the Nearest Integer) [. Use: R=0.083 L bar .mol-1 K-1]

Question

1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 × 10-3 bar The molar mass of the biopolymer is -104 gmol -1 (Round off to the Nearest Integer) [. Use: R=0.083 L bar .mol-1 K-1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

π=C R T π= osmotic pressure C= molarity T= Temperature of solution let the molar mass be M gm / mol 2.42 × 10-3 bar =(((1.46 g/M g m / mol))/0.1 l) ×((0.083 l-bar/mol-K)) ×(300 K ) ⇒ M=15.02 × 104 g / mol

Q. 1.46g of a biopolymer dissolved in a 100mL water at 300K exerted an osmotic pressure of 2.42×10−3 bar The molar mass of the biopolymer is________ −104gmol−1 (Round off to the Nearest Integer) [ Use : R=0.083L bar mol−1K−1]

π=CRT π= osmotic pressure C= molarity T= Temperature of solution let the molar mass be Mgm/mol 2.42×10−3 bar =0.1l(Mgm/mol1.46g​)​×(mol−K0.083l−bar​)×(300K) ⇒M=15.02×104g/mol

Q. $1.46 g$ of a biopolymer dissolved in a $100 m L$ water at $300 K$ exerted an osmotic pressure of $2.42 \times 1 0^{- 3}$ bar The molar mass of the biopolymer is________ $- 1 0^{4} g m o l^{- 1}$ (Round off to the Nearest Integer)
$[$ Use : $R = 0.083 L$ bar $m o l^{- 1} K^{- 1}]$