A solution is prepared by dissolving 5.88 g of an unknown nonelectrolyte in enough water to make 0.355 L of solution. The osmotic pressure of the solution is 1.21 atm at 27 °C. What is the molar mass of the solute? (R = 0.08206 L×atm/mol×K)
Answer
a.
30.3 g/mol
b.
0.00297 g/mol
c.
337 g/mol
d.
175 g/mol
e.
42.5 g/mol
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Answers & Comments
Verified answer
P = nRT/V
1.21 = n(0.082)(300) / 0.355L
n = 0.0175
5.88g / 0.0175 moles = 336 g/mole so c is the answer
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