$\,\therefore \,\,\,\frac{{{n_{NaHC{O_3}}}}}{{{n_{C{O_2}}}}}\,\, = \,\,\frac{2}{1}$
$\therefore \,\,{n_{NaHC{O_3}}}\, = \,2{n_{C{O_2}}}\,\, = \,\,2\, \times \,\frac{{112}}{{22400}}\, = \,0.01\,mole$
$\therefore \,\,{W_{NaHC{O_3}}}\, = \,0.01\, \times \,84\,\, = \,\,0.84\,g\,\,;\,$
${W_{N{a_2}C{O_3}}}\,\, = \,\,1.00\, - \,0.84\, = \,0.16g\,$
$\therefore \,\,\,\% \,N{a_2}C{O_3}\, = \,16$
વિભાગ $A$ |
વિભાગ $B$ |
1. 55.55 મોલ |
(P) સુક્રોઝના $6.022 \times10^{23}$ અણુ |
2. 2 મોલ |
(Q) 1.8 ગ્રામ $H_2O$ |
3. 0.1 મોલ |
(R) 126 ગ્રામ $HNO_3$ |
4. 0.01 મોલ |
(S) 1 લિટર શુદ્વ પાણી |