A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is g/mol.

160 g/mol

Explanation:

Step 1: Calculate the molarity of the solution

We will use the following expression.

π = M × R × T

where,

M = π / R × T

M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L

Step 2: Calculate the moles of solute in 550 mL (0.550 L)

0.550 L × 0.0752 mol/L = 0.0413 mol

Step 3: Calculate the molecular weight of the nonelectrolyte

0.0413 moles weigh 6.60 g.

6.60 g/0.0413 mol = 160 g/mol

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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