Question

Which of following solutions will have maximum boiling point elevation and which have minimum freezing point depression assuming the complete dissociation?
(a) 0.1 m KCl
(b) 0.05 m NaCl

(c) 1m AlPO4

(d) 0.1 m MgSO4​

Answer 1

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Solution : Boiling point elevation and freezing point depression are colligative properties that depend on number of particles in solution. The solution having more number of particles will have large boiling point elevation and that having less number of particles would show minimum freezing point depression.

(a) KCl K+ Cl

0.1m 0.1m 0.1m solution = 0.2 mol

(b) NaCl Na⊕ + Cl

0.05m 0.05m 0.05m solution = 0.1 mol

(c) AlPO4 Al3⊕ + PO4

1m 1m 1m solution = 2.0mol

(d) MgSO4 Mg2⊕ + SO4

0.1m 0.1m 0.1m solution = 0.2mole

AlPO4solution contains highest moles and hence highest number particles and in turn, the maximum ∆Tb NaCl solution has minimum moles and particles. It has minimum ∆Tf

Answer 2

Answer:

If alpha and beta are the zeros of the Quarditic polynomial p(x) = x2-x-4 then find 1/ alpha + 1/beta - alpha betaBrainly.in

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1

Secondary School Math 5 points

If alpha and beta are the zeroes of x2-x-4 find out the value of

1) 1/alpha + 1/beta - alpha beta

2) alpha/beta + beta/alpha +2 ( 1/alpha +1/beta) + 3*alpha*beta..

plzz its urgent......

Ask for details Follow Report by Arohi03 06.05.2018

Answers

THE BRAINLIEST ANSWER!

Sol : We have quadratic equation x² - x - 4.

Given α and ß are their zeroes.

We know that,

Sum of roots = - ( coefficient of x )/ coefficient of x²

α + ß = - ( - 1 ) / 1

α + ß = 1 / 1 = 1.

Now,

Product of roots = constant term / coefficient of x²

αß = ( - 4 ) / 1

αß = -4.

1. 1/α + 1/ß - αß

ß + α

= ------------- - αß

αß

By substituting the values of ( α + ß ) and ( αß ),

= ( 1 / -4 ) - ( - 4 )

= ( - 1 / 4 ) + 4

- 1 + 16

= ----------------

4

= 15 / 4.

2. α/ß + ß/α + 2 ( 1/α + 1/ß ) + 3αß

α² + ß² + 2ß + 2α

= --------------------------- + 3αß

αß

α² + ß² + 2 ( α+ß )

= ------------------------- + 3αß ----- eq.1

αß

Now ,we don't have the value of ( α² + ß² ), so let's find it ,

( α + ß )² = α² + ß² + 2 αß

By substituting the values of ( α + ß ) and αß in above equation,

( 1 )² = α² + ß² + 2 ( - 4 )

1 = α² + ß² - 8

α² + ß² = 1 + 8

α² + ß² = 9

Now by substituting the values of ( α² + ß² ) ,αß and ( α + ß ) in eq.1,

9 + 2 ( 1 )

= --------------- + 3 ( - 4 )

-4

9 + 2

= -------------- - 12

-4

- 11

= -------------- - 12

4

-11 - 48

= --------------

4

= -59/4.