There are 2 beakers, one with MgCl2 dissolved in water and one with NaCl dissolv

Question

There are 2 beakers, one with MgCl2 dissolved in water and one with NaCl dissolved in water. There is 1.0 L solutions of 1.0 M NaCl(aq) and 1.0 M MgCl2(aq). c. How many moles of NaCl and MgCl2 are in each beaker? d. How many moles of chloride ions in each beaker? How did you arrive at this answer? What is the concentration of chloride ions in each beaker? e. Explain how it is that the concentrations of chloride ions in these beakers are different even though the concentrations of each substance(compound) are the same. Say you were to dump out half of the MgCl2 solution from one beaker. a. What would be the concentration of the MgCl2(aq) ion and of the chloride ions in the remaining solution? b. How many moles of the MgCl2 of the chloride ions would remain in the beaker? c. Explain why the concentration of MgCl2 (aq) would change, whereas the number of moles of MgCl2 would change when solution was removed from the beaker. Consider the beaker containing 2.3 L of the 2.1 M NaCl(aq) solution. You now add 1.0 L of water of this beaker. a. What is the concentration of this NaCl(aq) solution? b. How many moles of NaCl are present in the 1.0 L of NaCl(aq) solution? c. Explain why the concentration of NaCl(aq) does change with the addition of water, whereas the number of moles does not change. Please explain how you came up with the answers.

Explanation / Answer

c)Inside the beaker NaCl exists as Na+ + Cl- ;

same for MgCl2 so for mol of NaCl(s) = 0 but of NaCl(aq) (which means in the form of ions)

n = Molarity*Volume = 1mol

same for MgCl2 = 1mol

d)NaCl --> Na+ + Cl-

hence 1 mol of NaCl gives 1 mol of Cl-

hence in 1st beaker 1 mol of Cl-

[Cl-] = 1/1=1M

MgCl2 --> Mg+ + 2Cl-

hence 1 mol of MgCl2 gives 2 mol of Cl-

hence in 2nd beaker 2 mol of Cl-

[Cl-] = 2/1=2M

e)Same as explained above since 1 mol of NaCl dissociates into only 1 mol of Cl- while 1 mol of MgCl2 dissociates into 2 mol of Cl-

(i)

a)Concentration doesn't change on sampling of a substance hence same [MgCl2] =1M

[Cl-] = 2M

b)moles of MgCl2 = 1*0.5 = 0.5mol

c)In the solutions formed the MgCl2 had homogenously distributed in the solution , hence when a part of the solution was taken a corresponding(proportional) no of moles of sbstance went out hence moles decreased in the ratio of volumes

since molarity is moles/volume it remained unaffected since both n and v decreased by the same proportion

(there was a typing mistake [] doesn't change)

ii)

a)n of NaCl = 2.1*2.3 = 4.83

[NaCl] = 4.83/(2.3+1) = 1.464M

b)in 1L n=1.464*1 = 1.464moles

c)Since the new water added did not contain any new moles of NaCl they remained the same as in initial solution since molarity = n/V and V increased hence Molarity decreased

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