Solved Chemistry Questions For University Freshers

 

Take the atomic weight of elements as follows C =12.011; H =1.00794; O =15.9994; N = 14.01.

Q1 (a) Your friend served you cup of tea. He told you that he used  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 0.024mol of table sugar (Sucrose C12H22011) in the tea he made. Calculate the mass of sugar he used in making the tea

Solution:

Q1(a). (Mole) n=Mass ​​ ​​ ​​ ​​ ​​​​ (C12H22O11=342gmol-1).

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ ​​ Mass

 : Mass =n×molar mass

 ​​ =0.024mol×342

   ​​ =8.208g

(b)i. If he makes choice to use aspartame C14H18O2N5​​ , an artificial sweetener ​​ instead, in his tea. The content of aspartame is 40mg (measured in two significant figures). Calculate how many moles of aspartame contained in the packet.

Solution (b)i

 Mass=40mg=40​​ ×10-3g of sweetener

 Molar mass (C14H18O2N5) =288gmol-1

(No of moles) n = massMolar Mass  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​   ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 

 ​​ ​​ ​​ ​​ ​​ ​​​​  ​​ =  ​​​​  40 ×10-3g288gmol-1​​ 

=  ​​ ​​ ​​​​ 0.0001388

 = 1.4×10-4​​ moles

 

 No of moles of Aspertame in the packet is 1.4×10-4​​ moles.

 

Q1b(ii) The mass of aspartame that would give 0.0120 mol is ? ​​ How many packets is this?

Solution Q1(b)ii

Mass = No of moles​​ ×​​ Molar mass

 0.0120​​ × 288

 ​​ ​​ ​​ ​​​​    = 3.456g

 1packet=40×10-3g

 

   Xpacket=3.456g

 

 Xpacket=86.4​​ 86 packets.

 

Q1c. From the equation below, derive the equation from first principle (half- life of radioactive materials)

t12​​ ----------------------------------------------(1)

 

 

 

Q1c.   It is a first order reaction

 

 

AB

 

​​  r = K[A]1---(1) Rate of reaction

 

 

 r1= d[A]--- (2)instantaneous rate

 

 

-d[A]dt​​ =k[A]  ​​ ​​ ​​ ​​​​ 

 

 ​​ ​​ ​​ ​​ ​​ ​​​​ 1[A]d[A] =-kdt (integrate both sides) [A]t

   ​​ 


 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 
A0At1 ​​ ​​​​ =  ​​​​ -​​ 0tdt ​​ ​​ ​​​​ 

   

ln[A]t​​ –ln[A]0​​ ​​ =-K(t-0)

 

 Kt = ln[A]0​​ –ln[A]t

 

Kt =ln[A]0  [A]t    

 ​​ ​​ ​​​​ 

 

(During half time t= t12)  ​​ ​​​​ [A]t​​ =12[A]0

 

Kt12​​ =ln[A]0

12[A]0

 

 Kt12 = ln2

 

 Kt12​​ =0.6932

 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ t​​ 12​​ = ​​​​ 0.6932k

 ​​ ​​ ​​​​ 

(Q1d) list and discuss any five (5) differences between nuclear reactions and chemical reactions​​ 

Solution Q1d.

Terms​​     Nuclear reaction  Chemical reactions​​ 

Energy

Released absorbed large amount of energy​​ 

​​ Released or absorbed small amount of energy

Tsotopes of the element​​ 

Show unidentical behavior​​ 

Show identical behaviour

Atoms

Nucleus of the atoms are affected​​ 

Only ​​ the electronic structures of the atoms are affected​​ 

Product​​ 

Results in conversion of one element to another​​ 

Does not result in the formation of new element ​​ 

Reactivity

Its not a function of its state of chemical combination​​ 

It’s a function of its state of ​​ chemical combination.

 

 

 

 

 

 

 

 

 

Q1e.Applications of radio isotopes

 

​​ Radio carbon dating – It is widely used by archaeologists to date articles composed of organic material e.g wood. Radio carbon dating of archaeological artifacts depends on the slow and constant production of radioactive C-14 in the upper atmosphere by neutron bombardment of nitrogen atoms

 

Food preservation-Foodstuffs are irradiated by a beam of​​ γ-rays which kill any bacteria present without imparting heat that could denature the food . However, there are concerns of food safety.

 

Blood flow monitoring-Sodium -24 (used as Nacl solution) is used to discover blood flow obstructions (blood clot) and other circulation disorders

 

Diagnosis/Treatment of thyroid disorders. Iodine -131 is used in the diagnosis ad treatment of disorders of the thyroid gland. Here, the patient is given low dose of the radioactive Iodine as Na. The radioactive iodine concentrates in the thyroid gland and activity measured.

 

 ​​​​ 

 

Q2a. Having far spent your feeding money, you considered ‘smoking garri’ with your 3 roommates by ‘soaking’ 960g of garri in a bowl with only 60 milligrams of sugar (sucrose, C12H22O11) left in the room, Assuming the garri was mixed properly with the sugar before water was added, calculate the concentration of sugar used in ppm (weight per weight on dry weight basis)

Q2a. Solution:​​ ​​ C​​ = ​​ mm.m ​​​​ × ​​​​ 1000Vcm3

 ​​ ​​ ​​ ​​ ​​​​ ​​ =  ​​ ​​ ​​ ​​ ​​​​ ​​ 60 ×103342 ​​​​ × ​​ 1000960

   ​​ ​​ ​​​​  

=  ​​ ​​ ​​ ​​ ​​ ​​​​ 0.000183

 =  ​​ ​​ ​​ ​​ ​​ ​​​​ 1.83×​​ 10-6

=  ​​​​ 183 ppm

 

(2b) An ancient wood artifact has a specific activity of 9.14d/min-g how old is the artifact? Assume the ratio of​​ 12C/14C in a living plant has specific activity of 15.3d/min-g

 

(2b) solution:  ​​​​ MM1​​ =​​ Q1-1Q

M = M1(15.3 -1)

M =9.14(15.3 - 1)

 ​​ ​​​​ =130.702

 ​​ ​​​​ =131yrs.

 

(2d) A bottle of wine of volume ​​ 750ml is labeled “ 10.5 % alcohol by volume , what is the volume of alcohol in the wine?

 

(2d) Solution: ​​ Volume of alcohol =​​   10.5100​​ × 7501

    ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ =78.75ml.

 

​​ 

(2e) The explosive trinitrotoluene (TNT) has the composition 37.1% C, 2.22% H, 18.50% N and ​​ 42.42% O ​​ what is the empirical formula of TNT?

 

(2e) Solution:  ​​ ​​​​ C   ​​ ​​ ​​​​ H   ​​ ​​ ​​​​ N   ​​ ​​​​ O

 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ ​​     37.0112 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ ​​ 2.221  ​​ ​​ ​​ ​​ ​​ ​​​​ ​​ 18.5014 ​​ ​​ ​​​​  ​​​​ 42.2716

  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​    ​​ ​​​​    ​​ ​​​​ 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 3.081.32  ​​ ​​ ​​ ​​ ​​ ​​​​ ​​ 2.221.32 ​​ ​​ ​​ ​​ ​​ ​​​​ ​​  1.321.32 ​​ ​​ ​​ ​​ ​​​​  ​​​​ 2.641.32

  ​​ ​​ ​​ ​​​​   ​​ ​​ ​​ ​​ ​​​​   ​​ ​​​​  

 

  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 2.33  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 1.68  ​​ ​​ ​​​​ 1   ​​ ​​ ​​​​ 2

 

T N T Empirical Formula C2H2N1O2

 

 

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