Answer
$5990.8\ g/mol$
Work Step by Step
Osmotic pressure:
$\Pi=cRT\rightarrow 3.1\ mmHg\times (1/760)atm/mmHg=c\times 0.082\ L.atm/mol.K\times (25+273)K$
$c=1.669\times10^{-4}\ mol/L$
Since the mass concentration is 1.00 g/L:
$M=c_m/c=1.00\ g/L\div 1.669\times10^{-4}\ mol/L=5990.8\ g/mol$
