Chemistry Some Basic Concepts of Chemistry

Mol wt A₂B = Mol wt of AB₃

 Hence (2A + B) = (A + 3B)

  $\therefore A=2B$

Atomic wt of A is two times of atomic wt of B.

$4HNO{I}_{3\left(\mathcal{l}\right)}+3KC{l}_{\left(s\right)}\underset{}{\overset{}{\to }}C{l}_{2\left(g\right)}+NOC{l}_{\left(g\right)}+2H_{2}O\left(g\right)+3KN{O}_3\left(g\right)$

$\downarrow$              

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$  

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$\therefore$  1 mole of KNO₃ produced by  $\frac{4}{3}$ moles of HNO₃

and  $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4*110}{3*101}$  moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole * mol.wt = 145 * 63 = 91.48 $\approx$ 91.5gm

No. of equivalents of H₂SO₄ = 100 * 0.1 * 2 = 20

No. of equivalents of NaOH = 50 * 0.1 = 5

No. equivalents of H₂SO₄ left = 20 – 5 = 15

150 * x = 15

$x=\frac{1}{10}=0.1N=1*10^{-1}N$

${\mathrm{ \; \; \; N}}_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

t =  0 56 $\mathcal{l}$            Excess 0

           -10 $\mathcal{l}$                     +20 $\mathcal{l}$

Final 46 $\mathcal{l}$                  20 $\mathcal{l}$

Vol. of N₂ (g) remained unreacted = 46 $\mathcal{l}$

% of C in organic compound

$=\frac{12}{14}*\frac{WC{O}_2}{W.O.C.}*100$

$=\frac{952.56}{21.648}=44\mathrm{\%}$

$\mathrm{\%}ofH=\frac{2}{18}*\frac{W{H}_{2}O}{W.O.C.}*100$

= $\frac{20}{18}*\frac{0.4428}{0.492}*100$ $\frac{}{}\frac{}{}$

= $\frac{88.56}{8.856}=10\mathrm{\%}$ $\frac{}{}$

= 100 – 54 = 46%

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$\therefore n_x=1$          

  $n_y=1$              

$n_z=\frac{0.05}{3}=0.0167$ here z is limiting reagent.

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$\therefore$ mass of xyz₃ = n * molecular mass

=  $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$

= 0.5 * 4 = 2g

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

= 0.00002 * 10³ mm

Molar mass of C₇H₅N₃O₆ = 12 * 7 + 1 * 5 + 14 * 3 + 16 * 6 = 84 + 42 + 96 = 227 g/mol

Number of moles = $\frac{681}{227}$  = 3 mol

$\therefore$ number of N-atoms

$=3*6.02*10^{23}*3=9*6.02*10^{23}$

$=5418*10^{21}$

$\therefore x=5418$ 

Number of moles of gas = $\frac{22.4}{22400}=\frac{1}{1000}moles$

Mass of nitrogen = $\left(\frac{1}{1000}*28\right)gm$

% of N in organic compound = $\frac{\left(\frac{28}{1000}\right)}{0.2}*100\mathrm{\%}=14\mathrm{\%}$

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

              Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

              Mass of C = $\frac{330}{44}*12$ gm = 90 gm

              Mass of H = $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$