Quantitative qs (prelim) (1 Viewer)

[stargaze]

1. The solubility of potassium nitrate is 85g per 100g of water at 50C and 20g per 100g at 10C. 60g potassium nitrate was dissolved in 100g at 50C. The solution was then slowly cooled to 10C. The amount of potassium nitrate, which crystallized out, was…

2. A solution of silver nitrate was made by dissolving 10.80g silver nitrate in water and making the volume to 250mL. The molarity of silver ions in this solution is...

3. The heat of solution of sulfuric acid ∆Hsoln kJ/mol. When 0.020 mol sulfuric acid is mixed with 100g water, the temp of the mixture will:

4. When 1g Magnesium Chloride dissolved in 200mL water is added to excess silver nitrate solution the number of moles of silver chloride foirmed is:

5. A person decided to take a bath using 250 L water at 46C. The quantity of heat needed to heat this amount of water from 16C to 46C is:


I cant seem to find examples of these types of qs in the book (conquering chem. Prelim). Would really appreciate it someone could show how to do such qs… Thanks a lot

 

[mervvyn]

1. answer is 40g --> only 20g can be dissolved at 10C, thus 60-20=40 will crystallise

2. molar weight of AgNO3 is 108 + 14 + 48 = 170g/mol (approx)
thus 10.80g is 10.80/170 = 0.0635 mol (approx)
molarity is moles over volume = 0.0635mol/0.25L = 0.254 mol (3sf)

3. i think something disappeared... but the dissolution of the acid will release 0.020*(heat of solution of H2SO4) kJ of energy = 'E' kJ
'E' = m*c*(change in temp)
= 100*4.18*T (where T is change in temp) and you could figure that out easily enough...

4. molar mass of MgCl2 is 24.3 + 35.5*2 = 95.3
thus no. of moles of MgCl2 is 1g/95.3 = 0.01 (approx)
as AgNO3 is excess, MgCl2 is limiting reagent

using an ionic equation: Ag+(aq) + Cl-(aq) = AgCl(s)
therefore no. moles Cl = no moles AgCl
there are 2*(no moles MgCl2) moles of Cl = 0.021 (approx, using unrounded result from 1st part)
therefore 0.021 mol of AgCl

5. Energy required = mass*specific heat*change in temp
= 250000g*4.18*30
= 31350 kJ

hope that helped.

 

[Numero Uno]

dunno about the rest
but
5 is
using
ΔH = - mCΔT
ΔH = - 25000g x 4.18 x 30
ΔH = - 313 5000 Joules
Since its - its an exothermic Reaction

 

[Numero Uno]

dunno about the rest
but
5 is
using
ΔH = - mCΔT
ΔH = - 25000g x 4.18 x 30
ΔH = - 313 5000 Joules
Since its - its an exothermic Reaction

 

[stargaze]

hmm thanks

I get those type of qs, thanks a lot .. Wondering if u could help me with a any of the following .. I promise u these will be the last!

1. How many moles of i) calcium ions ii) chloride ions are there in 17.7mL 0.0330 mol/L Calcium Chloride solution?

2. Find the molarity when 10mL 1.52mol/L HNO3 was diluted in 500mL

3. How much is needed of solution A is reach the concentration of solution B
(A = 0.143 mol/L , B = 0.0143 mol/L)
(between 250ml and 2L)

4. Tom easure the concentration of Mg ion in a solution an analyst took 50mL of the soln & added NaOH soln until no further ppt (of Mg(OH)2 ) was formed. The ppt was filtered off, dried and weighed; mass of 1.72g
a) The molarity of Mg ion in the original...
b) how many grams of Mg ion were there p/L in the original solution

5. Calculate the molar heat of soln for the ionic substance; 5.3g Calcium Chlroide was dissolved in 250mL water at 18.6. The temp rose to 22C.

6. 100mL of 0.022mol/L calcium chloride soln was mixed with 100mL 0.022 mol/L Sodium Sulfaet soln. The mixture was shaken but no ppt formed. Calculate concentration of Ca ions and SO4 ions in the mixed soln.

7. 1.2 x 10power-3 mercury (III) nitrate was dissolved in water and the volume made to 100mL. 5mL of this soln was diluted to 1L to form soln A.
a) molarity of soln A

 

[Xayma]

1. n_CaCl₂=n_Ca²⁺=1/2 n_Cl^-=0.0177*0.0330
n_Ca²⁺=0.0005841mol
n_Cl^-=0.0011682mol.

2. c=nv, n=c/v
n_initial=n_final
c_i/v_i*v_f=c_f
(Ill leave that to you to finish)

 

[stargaze]

umm.. xayma sorry but could u explain that???
q1 that is..

is that a formula or something?

 

[Xayma]

n=number of moles.

The number of moles of CaCl₂ is the same amount of moles as Ca²⁺ and 1/2 the number of Cl^- (since the 2 Cl atoms dissociate there are twice as many of them as there are CaCl₂ "molecules")