A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample.

Q- 9  A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

(i) express this in percent by mass

(ii) determine the molality of chloroform in the water sample.

Answer

part (i)

In this problem level is 15 ppm by mass

So   ppm is the part per millions

15 ppm =  mass of solute *10^6/mass of solvent    ………..(1)

Mass % CHCl3 = mass of solute*10^2 / mass of solvent  ...(2)

Divide equation second by first we get

 Mass %  of CHCl3  =  15* 10^-4 %

part 2 (ii)

Formula

 Molality = number of mole of solute / mass of solvent in Kg...(3)

 Here we have already find the mass % is 15*10^-4 

When even mass % is given take total mass of solution = 100 gram

And mass of solute (CHCl3)  will  =  15*10^-4  g

Mass of solvent = mass of solution – mass of solute

(Here mass of solute is very small as compare to total mass of solution so neglect it and we get)

Mass of solvent = mass of solution

Mass of solvent = 100 g

Now we have to find the number of moles of solute

 Number of moles of solute = mass of solute / molar mass …(4)

 So we have to calculate the molar mass of solute (CHCl3)

Molar mass of CHCl3                      =12+1+106.5

                                                         =119.5

Plug the value in equation (4)

No of moles of chloroform              = 15*10^4/119.5

                                                         =1.25*10^-5

Now plug the value of number of moles and mass of solvent in equation (4)

Here mass of solvent = 100 g

Convert it in Kg we divide by 100 we get

Mass of solvent = 0.1Kg 

Molality of                                          = 1.25*10^-5/0.1

                                                            = 1.25*10^-4 m

1.25*10^-4 m

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?

Answer

Step -1 (What is given in problem ?)

condition -1

Partial pressure of ethane = 1 bar

Mass of ethane =6.56 × 10–3 g
Condition -2
Now mass of ethane is 5.00 × 10–2
We have to find the pressure

So using Henry’s law 

Mass of dissolved gas(m) of directly proportional to its pressure (p)

m= kp   here k is proportionality constant

so plug the values we get

m1= kp1 ……………(1)

m2 = kp2 …………...(2)

Divide equation (2) by (1) equation we get

M2/m1 = p2/p1

Multiply by p1 both sides we get

P2 = m2*p1/m1

Plug the values now we get

                           P2 = 5.00 × 10–2* 1/6.56 × 10–3 g

                         P2= 7.62 bar (Answer)

State Henry’s law and mention some important applications?

State Henry’s law and mention some important applications?

Answer:-

Henry’s law :-

According to Henry’s law the solubility of gas in a liquid is directly proportional to the pressure of the gas

Mathematically

 mass of the dissolved gas  α  pressure of the gas

let mass of dissolved gas in unit volume            = m

pressure of gas                                                 = p

to remove sign of proportional we us a constant

here  KH is Henry’s law constant

so we get

m = K_H*p                            …………….(1)

we can measure mass of solute in term of molar fraction also

Hence our formula will

X = K_H*p                              …………. (2)

And we can write as

The partial pressure of the gas in vapour phase p is  proportional to the mole fraction of the gas x in the solution.

Importance of this Law :-

(1)     In Packing of soda cans :- To increase the solubility of CO2 gas in soda water , bottles of soda water is always packed under higher pressure. 

(2)     In Deep see diving: - As nitrogen is a more soluble gas in our blood and at deep see pressure increase so its solubility also increases, when scuba diver tries to come rapidly toward the surface of water, pressure decreased and Dissolved N2 gas comes back from the blood and make bubbles in his veins. It because bends .To avoid bends diver use oxygen diluted with helium because helium is less soluble in blood.

Q-11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

Q-11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer :-

When gas dissolve in liquid it generate a small quantity of heat so we can say that dissolution of gas in a liquid is a exothermic reaction.

Gas + liquid  < --- >  solution  + heat

 According to Le Chatelier's principle when we increase the temperature in a exothermic reaction , the reaction will move to back ward direction that means solution will dissociate and will give gas .Hence with the rise of temperature gas will less soluble.

Q-10 What role does the molecular interaction play in a solution of alcohol and water?

There is strong hydrogen bonding between the molecules of alcohol as well as in  the molecules of  water .when we mix both liquids, the intermolecular force of attraction decreases.

hydrogen bonding in alcohol and water

  Due to this decrees in intermolecular force , solution shows positive deviation from the ideal behavior. In the case of positive deviation vapor pressure increases and boiling point always decreases.

x mathematics conceptuals chapter -2

Degree of the polynomial :-  If p(x) is a polynomial in terms of  x, the highest power of x in p(x) is called the degree of the polynomial p(x).

For example,

 3x + 2 is a polynomial of  x.  Degree of expression is 1

 4z^2 –z + 2 is a polynomial of z.  Degree of expression is 2,

3x^3 – 2x – 2 is a polynomial of x.  Degree of expression is 3

Type of polynomials

linear polynomial :- A polynomial of degree 1 is called a linear polynomial.         

For example, 7x + 43,

quadratic polynomial:-  A polynomial of degree 2 is called a quadratic polynomial.

For example  x^2 + 3x + 7

cubic polynomial :- A polynomial of degree 3 is called a cubic polynomial

For example  x^3 + 3x

Zeros of the polynomial:-  A real number t  is called a zero of a polynomial if the value of f(t) = 0

For example  

f(x)  = x^2 – 6x +8

zeros of this equation are 2 and 4 because

f(2)= 2^2 -6*2 + 8 = 0

f(4)= 4^2 – 6*4 + 8 =0

Sum and product of root of quadratic equation   :-

For a equation ax^2 + bx  + c  = 0 , if  root are α and β ,

Roots for cubic equation :-

For a equation ax^3 + bx^2 + cx + d  = 0

Division Algorithm:- If p(x) and g(x) are any two polynomials with g(x) is not equal to 0, then we can find polynomials q(x) and r(x) such that

If r(x) = 0 or degree of r(x) < degree of g(x).

Dividend         = Divisor × Quotient   + Remainder

p(x)          = g(x)     × q(x)          + r(x),

X mathematics concept chapter -1

Euclid’s Division Lemma

For three positive integers a, b  there exists a unique integer q and r such that a = bq + r  and here value of  r will always less then b

That means if we divide number a by b and  q is our quotient and r is remainder then value of remainder will always less then deviser b .

For example 5 and 19

If we divide 22 by 5 we get 4 as quotient and 2 is remainder

So we can write it as  22= 5*4+ 2

And you can see that value of remainder is less then deviser

Composite number

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

The Fundamental Theorem of Arithmetic

If we ignore the order, any number which is more than 1, is either a prime number or can be written as unique product of prime number  .

Or

Every composite number can be written in term of product of unique set of prime numbers ignoring their

For example

26 is a composite number and it can be written as 2*13 according to fundamental theorem you cannot get another set of value which product is 26.

So   if we ignore the order of factors, prime factorization of all natural number is always unique.

Theorem of rational number: - for any prime number p which is divides a^2 then it will divide a also.

For example  5 divides 100 then it will divide root of 100( 10) also.

Theorem of terminator:-  for any rational number x which is written as p/q   and if q can be written in form of 2^n*p^m  and value of m and n are positive integer or equal to 0  then  decimal expansion of x will terminate.

For exampletake  x = 7/50

And 50 = 2*5*5 = 2^1*5^2

And powers 1 and 2 both are positive integer So value of x will terminate

Theorem of non-terminating repeating (recurring) :-  for any rational number x which is written as p/q   and if q ca not be  written in form of 2^n*p^m  and value of m and n are positive integer or equal to 0  then  decimal expansion of x will non-terminating repeating (recurring).

For example

X = 7/15

Here denominator 15 = 3*5

And it cannot be written in form of 2^n*5^m

x =0.4666……………

So x will non-terminating repeating (recurring) .