TS EAMCET 2019 :: Chemistry :: General questions

• Chemistry - General questions

Find the reagent that oxidises glucose into saccharic acid?

Answer:

Explanation:

On oxidation  with $HNO_{3}$ AND $H_{2}O$, the glucose give saccharic acid as follows:

 Hence, option (c) is the correct answer.

The products formed during the reaction of carbon with conc. $H_{2}SO_{4}$ are

$C+ H_{2}SO_{4} $           $\underrightarrow{ \triangle}$  products

Answer:

Explanation:

 when $H_{2}SO_{4}$   (conc.)  react with non-metals, it give oxoacid of non-metal and $SO_{2}$ (g)  . If oxoacid is unstable , it decomposes.[as in $H_{2}CO_{3}$, which decompose as $CO_{2}+H_{2}O]$ the whole reaction occurs as follows:

    $C+H_{2}SO_{4}$   $\underrightarrow{\triangle}$   $ H_{2}CO_{3}+SO_{2} \rightarrow CO_{2}+H_{2}O+SO_{2}$

Hence , option (b) is  the correct answer

The first-order decomposition of $H_{2}O_{2}$ in an appropriate medium is characterised by a rate constant of 0.2303 min^-1 . What is the time  (in min) required to complete 9/10 fraction of the reaction?

Answer:

Explanation:

Given, rate constant (k) for 

first order=0.2303 min^-1

 For first  order reaction,

 Time ,$(t)= \frac{2.303}{k}\log \left[\frac{a}{a-x}\right]$  ...........(i)

 where, a= initial  concentration

(a-x)= concentration at time (t)

 x= amount complete

 let, initial concentration (a)=1

                               x=$\frac{9}{10}$

 then , (a-x)=$1-\frac{9}{10}=\frac{1}{10}$

 From equation (i),

$t= \frac{2.303}{0.2303}\log\frac{1}{\left(\frac{1}{10}\right)}$

  t= 10 log 10

 $t= 10 \times 1=10 min^{-1}$

 Hence, option (b) is correct

A  camphor  sample melts at $176 ^{0} C.$ $K_{f}$ for  camphor is 40 K kg mol^-1 . A solution of 0.02g  of a hydrocarbon  in 0.8 g of camphor melts at $156.77^{0}C$ . Hydrocarbon is  made up of  92.3% of carbon. What is the molecular formula of the hydrocarbon?

Answer:

Explanation:

Melting point of camphor= $176 ^{0}$ C

$ (K_{f})$ for camphor=40 K kg mol^-1

 Mass of hydrocarbon (w_B)=0.02 g

Mass of camphor (w_A)=0.8 g

 Temperature of camphor (at which it melts)= $156.77^{0} C$

 Thus, $\triangle T_{f}$ for  camphor =176-156-77=   $19.23^{0} C$=19.23 K

 $\therefore$ depression in freezing point ,

     $\triangle T_{f}$ =  $\frac{k_{f}\times w_{B}}{M_{B}}\times\frac{1000}{w_{A}}$

              $19.23= \frac{40\times0.02\times1000}{M_{B}\times0.8}$

 Molar mass of solute (M_B)

=$\frac{40\times0.02\times1000}{19.23\times0.8}=52$

 $\therefore$   Empricial formula = CH

 Empirical  formula  weight =12+1=13

 Multiple (n) = Molecular weight/ Empirical  formula weight = $\frac{52}{13}=4$

 Thus, molecular formula = (CH)₄= C₄H₄

The vapour pressure of pure $CCl_{4}$ (molar  mass=154 g mol^-1) and $SnCl_{4}$ (molar mass=170 g mol^-1) at $25^{0} C$ are 115.0 and 238.0  torr respectively . Assuming ideal behaviour  , calculate the total approximate vapour  pressure  in torr of a solution containing 10 g of $CCl_{4}$  and 15 g of $SnCl_{4}$.

Answer:

Explanation:

Given, vapour pressure of pure $CCl_{4}$ at 25° C

           $(P^{0}_{CCl_{4}})=$  115.0 torr

 Vapour pressure  of pure $SnCl_{4}$ at $25^{0}$C

                 $(P^{0}_{SnCl_{4}})=$  238.0 torr

 Molar mass of $CCl_{4}$   $ [M_{(CCl_{4})}]=154 g$$mol ^{-1}$

Molar mass of $SnCl_{4}$  in solution  $[M_{SnCl_{4}}]$= 170 g $mol^{-1}$

 Mass of $CCl_{4}$ in solution  $(w_{CCl_{4}})=10g$

 mass of $SnCl_{4}$ in solution    $(w_{SnCl_{4}})=15 g$

 from Dalton's law of partial pressure,

$p_{total}=\chi_{CCl_{4}}.p^{0}_{CCl_{4}}+\chi_{SnCl_{4}}. p^{0}_{SnCl_{4}}$

 and     $\chi=\frac{n_{simple}}{n_{total}}=\frac{w}{M \times n_{total}}$

Now,  $n=\left[\frac{w}{M}\right]_{CCl_{4}}=\frac{10}{154}=0.065$

and     $\left[\frac{w}{M}\right]_{SnCl_{4}}=\frac{15}{170}=0.09$

  Thus,    $\chi_{CCl_{4}}=\frac{n_{CCl_{4}}}{n_{CCl_{4}}+n_{SnCl_{4}}}$

                     = $\frac{0.065}{0.06+0.09}$

 $\chi_{CCl_{4}}=\frac{0.065}{0.15}=0.43$

 Therefore,   $\chi_{SnCl_{4}}=1-0.43=0.57$

 Thus,     $p_{total}$= $0.43 \times 115.00+0.57 \times 238$

  =49.45+135.66=  185.11≈185.85 torr

 Hence, option (a) is the correct answer 

• Chemistry - General questions