calculate the number of moles of solute present in each of the following solutions?
1. calculate the number of moles of solute present in 270 ml of 1.40 m hno3(aq).

2. calculate the number of moles of solute present in 55.0 mg of an aqueous solution that is 1.65 m nacl. assume that for dilute aqueous solutions, the mass of the solvent is the mass of solution.

3. calculate the number of moles of solute present in 75.0 g of an aqueous solution that is 1.40 % sucrose (c12h22o11) by mass.
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
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. 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 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** 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 answer: the molecular mass of an element is defined as the sum of the masses of the elements present in the molecule. molecular mass is obtained by multiplying the atomic mass of an element with the number of atoms in the molecule and then adding the masses of all the elements in the molecule. o2 is a diatomic molecule. the mass of an oxygen atom = 16 amu. ∴ mass of o2 molecule = 2 × 16 = 32 amu. mass of one molecule of oxygen → 32/ avogadro constant ( 6.023 × 1023) → 5.31 × 10-23g. 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) setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v we’re being asked to determine which of the given solutions is a good buffer system.recall that a buffer can be composed of the following:• weak base and its conjugate acid• weak acid and its conjugate base• weak acid and a strong base ▪ concentration of weak acid > concentration of strong base• weak base and a strong acid ▪ concentration of weak base > concentration of strong acid[readmore]a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.potassium hydroxide → koh → group 1a k+ + oh- → strong basepotassium bromide → kbr breaks up in the solution: kbr → k+ + br-▪ k+ is a neutral ion → will not contribute to ph  hbr       +      h2o       →      br-      +         h3o+(weak acid)                            (conjugate base)•  strong base + weak base• not a good buffer systemb. 0.28 m hydroiodic acid + 0.18 m sodium iodide. hydroiodic acid → hi → weak acidsodium iodide → nai → breaks up in the solution: nai → na+ + i-hi       +      h2o       →      i-      +         h3o+(weak acid)                            (conjugate base)• weak acid + conjugate base = buffer• the solution is a good buffer systemc. 0.30 m ammonia + 0.37 m ammonium nitrate. ammonia → nh3 → weak baseammonium nitrate → nh4no3 → breaks up in the solution: nh4no3 → nh4+ + no3-nh3       +      h2o       →      nh4+      +         oh-(weak base)                            (conjugate acid)• weak base + conjugate acid = buffer• the solution is a good buffer systemd. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid. hypochlorous acid → hclo → weak acidhydrobromic acid → hbr → weak acid•  weak acid + weak acid• not a good buffer systeme. 0.38 m calcium bromide + 0.30 m barium bromide.calcium bromide → cabr → breaks up in the solution: cabr → ca+ + br-barium bromide → babr → breaks up in the solution: babr → ba+ + br-•  conjugate base + conjugate base• not a good buffer systemanswer: the following aqueous solutions are good buffer systems b. 0.28 m hydroiodic acid + 0.18 m sodium iodide and c. 0.30 m ammonia + 0.37 m ammonium nitrate