Chemistry Some Basic Concepts of Chemistry

% of C =   $\mathrm{\%}ofC=\frac{WtofC{O}_2}{Wtofcomp.}*\frac{12}{44}*100=\frac{2.64}{1.8}*\frac{12}{44}*100=40\mathrm{\%}$

$\mathrm{\%}ofH=\frac{Wtof{H}_{2}O}{Wtofcomp.}*\frac{2}{18}*100=\frac{1.08}{1.8}*\frac{2}{18}*100=6.67\mathrm{\%}$       

% of O = 100 – (40 + 6.67) = 53.33%

NaOH + Na₂CO₃

(1) When Hph is added

m.e NaOH + $\frac{1}{2}m.eN{a}_{2}C{O}_3=m.e.HCl=17.5*\frac{1}{10}=1.75$  (m.e = milli equivalents)

(2) When MeOH is added after Hph

$\frac{1}{2}meN{a}_{2}C{O}_3=m.eHCl=1.5*\frac{1}{10}=0.15$

$\begin{array}{l}\therefore m.eNaOH=1.75-0.15=1.6\\ m.eN{a}_{2}C{O}_3=0.15*2=0.3\end{array}$  

$\frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}*1000=0.3$   

$Weight\mathrm{\%}ofN{a}_{2}C{O}_3=\left(\frac{\frac{0.3*53}{1000}}{0.4}\right)*100=\frac{0.3*53}{10*0.4}=\frac{15.9}{4}=3.975\mathrm{\%}\approx 4\mathrm{\%}.$               

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

Total wt = 4gm = W_NaOH + $W_{N{a}_{2}C{O}_3}$  

Let us suppose moles are 'm' for each.

4 = 40m + 106m

$\therefore m=\left(\frac{4}{146}\right)$ moles of NaOH and Na₂CO₃ each .

$\therefore$ Mass of NaOH (x gm) = $\frac{4}{146}*40=1.095\approx 1.0$

Molarity (M) = $\frac{w*1000}{molecularmass*volumeofsolution\left(ml\right)}$

= $\frac{6.3*1000}{126*250}=\frac{4}{20}=0.2M$

Molecular mass of oxalic acid $\left(H_2{C}_2{O}_4.2H_{2}O\right)$

= 1 * 2 + 12 * 2 + 16 * 4 + 2 * 18

              = 26 + 64 + 36 = 126

              M = 2 * 10^-1 M

              = 20 * 10^-2M

$\therefore x*10^{-2}=20*10^{-2}$

$\therefore x=20$

Ans. = 20

23 * 3 gm of Na contained by 164 gm of Na₃PO₄

So, 3.45 gm of Na contained by $\left(\frac{164*3.45}{69}\right)$ gm of Na₃PO₄

Therefore mole of Na₃PO₄= $\frac{164*3.45}{69*164}$ = 0.05 mol

Molarity = $\frac{0.05*1000}{100}$ M = 0.5 M

= 50.0 * 10^-2 M

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

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