A solution is created by dissolving 11.0 grams of ammonium chloride in enough wa

Question

A solution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 345 mL of solution. How many moles of ammonium chloride are present in the resulting solution?

When thinking about the amount of solute present in a solution, chemists report the concentration or molarity of the solution. Molarity is calculated as moles of solute per liter of solution. What is the molarity of the solution described above?

to carry out a particular reaction, you determine that you need 0.0500 moles of ammonium chloride. What volume of the solution described above will you need to complete the reaction without any leftover NH4Cl?

Please explain how you did all of this because I'm having a really hard time understanding all this stuff

Explanation / Answer

Number of moles of NH4Cl = ( mass of NH4Cl ) / ( Molar mass of NH4Cl)

= ( 11 g) / ( 53.49g/mol) = 0.206 mol ( molar mass can be taken from web or calculated by adding atomic masses of N + 4H + Cl )

2) Molarity of NH4Cl solution = ( moles of Nh4Cl) / ( solution volume in L)

= ( 0.206 mol) / ( 0.345L) = 0.596 mol/L or 0.596 M

3) we need 0.05 moles of NH4Cl , we use formula to calculate volume

Molarity = moles / volume

0.596 mol/L = 0.05 moles / ( volume)

volume = ( 0.05mol) / ( 0.596mol/L) = 0.0839L = 0.0839 x 1000 ml = 83.9 ml

Thus volume needed = 83.9 ml

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