please help with these two questions What is the final concentration of chloride

Question

please help with these two questions

What is the final concentration of chloride ion, [Cl^-], in a solution made by mixing 158 ml of 0.200 M NaCl with 200 mL of 0.150 M MgCl_2? How many milliliters of a 0.400 M NaCl solution would you need to add to 1.00 L of a 2.0 Times 10^-2 M AgNO_3 solution to completely precipitate all of the silver as AgCl(s)? How many prams of AgCl would precipitate by mixing 50 mL of 0.4 M NaCl with 80 mL of 0.2 M AgNO_3? You'll need to determine the limiting reactant first. FWT of AgCl = 143.31 g/mol

Explanation / Answer

Moles of NaCl in 158ml of 0.2M= Molarity * Volume in L

=0.2*158/1000=0.0316 moles

NaCl -à Na+ Cl-

So moles of Cl- are same as that of NaCl

Hence moles of Cl- = 0.0316

Moles of MgCl2 in 200ml of 0.15M =0.15*200/300 =0.03 moles

MgCl2 -à Mg+2 + 2Cl-

Moles of Chloride ions = 2* 0.03= 0.06

Total moles of Cl- =0.0316+0.06= 0.0916

Total volume after mixing =200+158= 358 ml= 0.358 L

Molarity of Cl- =0.0916/0.358 moles/L=0.25886 M ( E is the correct answer)

B.

Moles of AgNO3 in 1L of 2*10-2 AgNO3 =2*10-2 moles

The reaction is AgNO3+ NaCl--à AgCl+ NaNO3

1 mole of AgNO3 require 1 moles of NaCl for precipitating as AgCl.

2*10-2 Moles of A require 2*10-2 moles of NaCl

But 0.4 moles of NaCl is there in 1 L of solution

2*10-2moles of NaCl is there in 2*10-2/0.4 L= 2*10-2*1000/0.4 ml=50 ml ( B is the correct answer)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.