A solution is made by dissolving 37.5g of sodium sulfide (Na2S) in

217g of water.

What is the molality of the solution?

THE MOLARITY IS 2.22 MOL/DM3

Explanation:

The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water

Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol

Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;

37.5 g of Na2S = 217 g of water

( 37.5 * 1000 / 217 ) g = 1000 g of water

So, 172.81 g/dm3 of the solution

So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass

Molarity = 172.81 g/dm3 / 78 g/mol

Molarity = 2.22 mol/dm3

The molarity of the solution is 2.22 mol/dm3

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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