Chemistry Some Basic Concepts of Chemistry

Moles of $\mathrm{H}\mathrm{C}\mathrm{l}=$ moles of ${\mathrm{N}\mathrm{H}}_3$

$=2*$ moles of urea $=2*\frac{0.6}{60}=0.02$

$\Rightarrow \mathrm{N}.\mathrm{V}=0.02$

Theory based.                                                                                                                                                                 

50000.020 * 10^-3

The significant figure in the given number is 8.

% 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