Answer to Question #21605 in Organic Chemistry for Sadaam Ahmed




 A sample of a substance known to contain chloride ions was dissolved in distilled water in a 1L volumetric flask. Then 25.00 mL of this solution was treated with excess silver nitrate, AgNO3 (aq). The precipitate of silver chloride, AgCl (s) was filtered and dried. The mass of the dry precipitate was 0.765 g. 
 
 a. Calculate the concentration of chloride ions.
 b. If the original substance was sodium chloride, NaCl (s), what mass of it was dissolved in the volumetric flask?
AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g**
a) ans. concentration = no. of moles of solute * 1000/ vol. in m;

no. of moles of solute = concentration * vol / 1000 = 1.4 * 270 / 1000 = 0.378 = 0.378 mol.



b)

molality = 1.65

molality = moles of solute / mass of solvent in kg

1.65 = moles of solute / 5.5 x 10^-5 kg

moles of solute = 9.075 x 10^-5

c)

mass of sucrose = 75 x 1.4 / 100 = 1.05g

moles = 1.05 / 342 = 3.07 x 10^-3 mass (molar mass) → mol (avogadro's number) → molecules[readmore]3.51×104 g ×1 mol163.94 g×6.022×1023 molecules1 mol = 1.29x1026 moleculesthere are 1.29x1026 molecules in the sample recall that at the equivalence point of a titration:[readmore]moles acid=moles basealso, recall that moles = molarity × volume. solving for molarity of naoh:0.568 g 1 mol khp204 g1 mol oh-1 mol khp1 mol naoh1 mol oh-136.78 ml1 ml1x10-3 l = 0.0757 mollor mtherefore, the molarity of naoh needed is 0.0757 m kclo3(s) — heat→ kcl (s) + o2(g)      reactant        productk        1                   1cl       1                   1o       3                    2[readmore]2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g)      reactant        productk        2                   2cl       2                   2o       6                    6the balanced equation for the reaction is 2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g) hey there! as answered previously:-
crystallization is the process of slow precipitation of solid
crystals of a substance from a concentrated homogeneous solution it
is basically a separation method in which a substance is
transferred from the liquid (i e solution) phase to the solid state
the substance is thus obtained in pure crystalline form and has no
impurity it is used by chemists for the purification of solid
substances
following is an example of using crystallization to obtain pure
copper sulphate from an impure sample
1) take a small amount (5 g) of an impure sample of copper
sulphate in a china dish
2) dissolve the sample in 10 ml of water (we basically have to
use minimum amount of water in order to prepare a saturated
solution)
3) then, filter it to remove impurities
4) evaporate the filtrate on a water bath till small crystals
are formed indicating that a concentrated solution has been
obtained
5) after this, cover the solution with a filter paper and leave
it undisturbed at room temperature for the rest of the day
![](https://img-nm%20mnimgs%20com/img/study_content/lp/1/9/8/114/318/760/662/5-6-08_sonali_9_chemistry_9%202%202%202%201_gsx_sg_ss_html_3c9d32fa%20png)
it will be observed that crystals of pure copper sulphate are
obtained and the impurities are left behind in the solution for water, h2o1 mol h2o = 18.0 g h2o1 mole h2o = 6.022x1023 h2o molecules[readmore]the number of h2o molecules in a 9.00-g sample of water:9.00 g h2o×1 mol h2o18.0 g h2o×6.022×1023 h2o molecules1 mol h2o= 3.01x1023 h2o moleculesthe number of h2o molecules in a 9.00-g sample of water is 3.01x1023 h2o molecules. NaBr   

  

SO3   

  

CaCl2   

  

Mg3N2 a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 since the solution is a buffer, we can use the henderson-hasselbalch equation to calculate the ratio of [so32-]/[hso3-] to have a ph of 7.39:ph=pka+logconjugate baseweak acidin the solution, identify the weak acid and the conjugate base. recall from the bronsted-lowry definition that an acid is a proton (h+) donor and the base is a proton (h+) acceptor.• nahso3 is a salt that will give hso3- when it dissociatesnahso3 → na+ + hso3-• na2so3 is a salt that will give so32- when it dissociatesna2so3 → 2 na+ + so32-•hso3- is a weak acid.          - concentration of hso3- = concentration of nahso3• when hso3- loses h+ to give a conjugate base, so32- is formedhso32- - h+ → so32-- concentration of so32- = concentration of na2so3      hso3-    +   h2o     →        so32-        +       h3o+(weak acid)    (base)      (conjugate base)  (conjugate acid)we can rewrite the henderson-hasselbalch equation as:ph=pka+logso32-hso3-calculate [so32-]/[hso3-] ratio:• ph = 7.39• pka = 7.19ph=pka+logso32-hso3-7.39=7.19+logso32-hso3-logso32-hso3-=7.39-7.19logso32-hso3-log=0.2loglog has a base of 10 and when you divide a number by log you will take 10 to the power of the numberso32-hso3-=100.2[so32-]/[hso3-] = 1.58the ratio [so32-]/[hso3-] is 1.58 **Pure substance – ethyl alcohol, oxygen, carbon, distilled water mixture – blood, steel**