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Question: A solution of \({\text{XY}}\) \({\text{100}}\% \) ionised has osmotic pressure equal to four times t...

A solution of XY{\text{XY}} 100%{\text{100}}\% ionised has osmotic pressure equal to four times the osmotic pressure of 0⋅01M BaCl20 \cdot 01M{\text{ }}BaC{l_2}(100%{\text{100}}\% ionised). Find the molarity of XY{\text{XY}}.
A) 6×10−2M6 \times {10^{ - 2}}M
B) 3×10−2M3 \times {10^{ - 2}}M
C) 4×10−2M4 \times {10^{ - 2}}M
D) 12×10−2M12 \times {10^{ - 2}}M

Explanation

Solution

In the given question the unknown compound XY{\text{XY}} and BaCl2BaC{l_2} both are 100%{\text{100}}\% ionised which means one can compare them by using the relevant equation. One can use the formula of the osmotic pressure to find out the value of the Van’t Hoff factor and evaluate this value into the further equations to find out the molarity.

Complete step by step answer:

  1. First of all let us analyze the facts given in the question where the osmotic pressure of XY{\text{XY}} is equal to the four times the osmotic pressure of BaCl2BaC{l_2} and we can compare them by using the osmotic pressure formula. Let us represent this statement in a mathematical formula,
    The osmotic pressure of XY{\text{XY}}  = {\text{ = }} 4×{\text{4}} \times Osmotic pressure of BaCl2BaC{l_2}
  2. Now let's write the osmotic pressure formula,
    π=iCRT\pi = iCRT
    Where, i=i = Van’t Hoff factor, π=\pi = Osmotic pressure, C = {\text{C = }} the concentration of a solution, R = {\text{R = }} Gas constant == 0⋅0823 L⋅atm⋅K−1⋅mol−10 \cdot 0823{\text{ L}} \cdot {\text{atm}} \cdot {{\text{K}}^{ - 1}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}, T=T = Temperature in kelvin
  3. Now we don’t know the value of Van’t Hoff factor and that value is nothing but the number of moles or ions of the product. So let's write the general reaction for this reaction for XY{\text{XY}},
    XY→X+ + Y−{\text{XY}} \to {{\text{X}}^ + }{\text{ + }}{{\text{Y}}^ - }
    The number of ions formed is two hence, van’t Hoff factor value of XY{\text{XY}} (i)=2\left( i \right) = 2
  4. Same way the general reaction for the reaction BaCl2BaC{l_2},
    BaCl2→Ba2+ + 2Cl−BaC{l_2} \to B{a^{2 + }}{\text{ + 2C}}{{\text{l}}^ - }
    The number of ions formed is three hence, van’t Hoff factor value of BaCl2BaC{l_2} (i)=2\left( i \right) = 2
  5. As both the compounds get 100%{\text{100}}\% ionised and osmotic pressure of XY{\text{XY}} equal to four times the osmotic pressure of BaCl2BaC{l_2} we can say that,
    π(XY)=4π(BaCl2)\pi \left( {XY} \right) = 4\pi \left( {BaC{l_2}} \right)
    As the values in the osmotic pressure formula RT{\text{RT}} are constant for both sides, we can take other factors in the above equation,
    i×[XY]=4×i×[BaCl2]i \times \left[ {XY} \right] = 4 \times i \times \left[ {BaC{l_2}} \right]
    =>2×[XY]=4×3×[BaCl2]2 \times \left[ {XY} \right] = 4 \times 3 \times \left[ {BaC{l_2}} \right]
    As the concentration of BaCl2BaC{l_2} is given as 0⋅01 M0 \cdot 01{\text{ M}} ,
    2×[XY]=4×3×[0⋅01]2 \times \left[ {XY} \right] = 4 \times 3 \times \left[ {0 \cdot 01} \right]
    =>2×[XY]=0⋅122 \times \left[ {XY} \right] = 0 \cdot 12
    Now lats take the value of XY{\text{XY}} on one side,
    [XY]=0⋅122=0⋅06 M\left[ {XY} \right] = \dfrac{{0 \cdot 12}}{2} = 0 \cdot 06{\text{ M}}
    Hence we got the value of the concentration of XY{\text{XY}} as 0⋅06 M0 \cdot 06{\text{ M}} which also can be written as 6×10−2 M6 \times {10^{ - 2}}{\text{ M}} which shows option A as a correct choice.

Note:
When a compound is 100%{\text{100}}\% ionised it means it is completely ionised where the complete ionisation is complete separation of positive and negative charged parts present in that molecule. The positive and negative ionic particles act as individual particles after the complete ionisation of the molecule.