MHT CET 2019 :: Chemistry :: General questions

• Chemistry - General questions

If $C(s)+O_{2}(g)\rightarrow CO_{2}(g);$  $\triangle H_{}=-X$     

$CO(g)+\frac{1}{2}O_{2}(g)\rightarrow CO_{2}(g);$   $\triangle H_{}=-Y$  .

Calculate  $\triangle _{f}H$ for CO (g) formation

Answer:

Explanation:

$C(s)+O_{2}(g)\rightarrow CO_{2}(g);\triangle H_{1}=-x$     ...........(i)

$CO(g)+\frac{1}{2}O_{2}(g)\rightarrow CO_{2}(g);\triangle H_{2}=-y$  ...........(ii)

 For the formation of CO subtract Eqs.(ii) from (i)  i.e,

  $C(s)+O_{2}(g)\rightarrow CO_{2}(g)$

$\underline{CO(g)_{(-)}+\frac{1}{2}O_{2}(g)_{(-)}\rightarrow CO_{2}(g)_{(-)}}$

$\underline{C(s)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)}$

 $\therefore$    $\triangle_{r}H$ for formation of CO=  $\triangle H _{1}-\triangle H _{2}$

                                                                    =-x+y   or y-x

9-gram anhydrous oxalic acid (mol.wt.=90) was dissolved in 9.9 moles of water. If the vapour  pressure of pure water is p₁° the vapour  pressure of the solution is 

Answer:

Explanation:

  The total vapour pressure of a solution in this case only depends on vapour pressure of water  as anhydrous oxalic acid  is a non-volatile compound.

 $\therefore$  Vapour pressure of solution= vapour pressure of  water (p_w)

According  to Raoult's law= 

 $p_{w}=x_{w}p_w^o$

 Number of moles  of oxalic acid = $\frac{9}{90}$= 0.1 moles

$\therefore$    $x_{w}=\frac{9.9}{9.9+0.1}=0.99$

 $\Rightarrow $   $p_{s}=p_{w}=$    $0.99p_1^0$

 The ionic charges of manganate and permanganate ion are respectively.

Answer:

Explanation:

 The chemical formula for manganate and permanganate ion respectively are MnO₄^2-  and MnO₄^-. Thus, the ionic charge of manganate and permanganate ion are respectively  -2  and -1

Relationship between van't Hoff 's factor (i) and degree of dissociation ($\alpha$ ) is 

Answer:

Explanation:

 Relationship between vant Hoff factor (i) and degree of dissociation  ($\alpha$) is given by

$\alpha=\frac{1-i}{n'-1}$

 where, n is the number of ions formed after dissociation .

 The relationship  can be obtained  as follows:

 For the reaction,   $A\rightleftharpoons n'B$

initially               1 mole      0

 after dissociation  (1- $\alpha$) mole n' $\alpha$

 Total number of moles present  in the solution 

   = $(1-\alpha)+n'\alpha=1+(n'-1)\alpha$

 van't Hoff factor ,   $i=1+(n'-1), \alpha >1 if n'\geq2$

 $\therefore$     $\alpha=\frac{i-1}{n'-1}$

 Isobutylene  on hydroboration followed  by oxidation  with hydrogen peroxide  in presence of base yields

Answer:

Explanation:

 Isobutylene on hydroboration followed by oxidation with hydrogen peroxide in presence of base yields isobutyl alcohol.

Note: The alcohol  formed is in accordance with anti-Markovnikov's rule 

• Chemistry - General questions