An experiment procedure starts with 625.0 mL of a 9.10 M hydrocyanic acid (HCN(aq)) solution. During...

An experiment procedure starts with 625.0 mL of a 9.10 M hydrocyanic acid (HCN(aq)) solution. During the course of the experiment, the volume of the this acid solution is increased by 1.457 L by the addition of water solvent as part of the outcome. Based on this information, what would be the altered molar concentration of the aqueous HCN solution?

Expert Solution

Given data: V = 625 ml = 0.625 L, Molarity = 9.10 M

Moles of HCN = Molarity × Volume

= 9.1 × 0.625 = 5.6875 moles.

Now 1.457 L of water is addedto the HCN. Then

V = 0.625 + 1.457 = 2.082 L, Mol of HCN = 5.6875 moles

Concentration of HCN = Moles of HCN in solution / Volume of Solution

= 5.6875 moles / 2.082 L = 2.73 M.

Hence the final concentration of aqueous HCN solution is 2.73 M.

queen_honey_blossom answered 2 years ago

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