How many moles of solute are present in a sodium chloride solution with a total volume of 1.25 L if the solutions known concentration is 0.330 M?
3.788 mol NaCl
0.264 mol NaCl
1.58 mol NaCl
24.14 mol NaCl
0.413 mol Naci

n = 0.413moles (3 sig-figs).

Explanation:

From definition of molarity, solve for moles and substitute given data.

Molarity(M) = Moles Solute(n)/Volume of Solution in Liters(V)

M = n/V => n = M·V => 0.330moles/Liter x 1.25Liters = 0.4125moles ≅ 0.413moles (3 sig-figs).

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

24