# Chapter 13 Solutions and Their Behavior - Study Questions - Page 505b: 39

$5990.8\ g/mol$

#### 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: $M=c_m/c=1.00\ g/L\div 1.669\times10^{-4}\ mol/L=5990.8\ g/mol$

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