Question

Question 32
Fill in the missing data in the Table 3.1
$\begin{array}{ccccc}Speciesproperty& H_{2}O& C{O}_2& Na-atom& MgC{l}_2\\ Numberofmoles& 2& -& -& 0.5\\ Numberofparticles& -& 3.011\times {10}^{23}& -& -\\ Mass& 36g& -& 115g& -\end{array}$

Answer 1

For $H_{2}O$
Given number of moles = 2 and mass = 36 g
$\therefore$ Number of particles = number of moles $\times 6.022\times {10}^{23}$
$=2\times 6.022\times {10}^{23}=1.2044\times {10}^{24}$
For $C{O}_2$
Given, number of particles $=3.011\times {10}^{23}$
$\therefore$ Number of moles of $C{O}_2=\frac{numberofparticles}{6.022\times {10}^{23}}$.
$=\frac{3.011\times {10}^{23}}{6.022\times {10}^{23}}=0.5mol$
Mass of $C{O}_2$ = moles $\times$ molar mass
$=0.5\times 44=22g$
For Na-atom
Given, mass = 115 g
Number of moles $\frac{mass}{molarmass}=\frac{115}{23}=5mol$
$\therefore$ Number of particles $=5\times 6.022\times {10}^{23}=3.011\times {10}^{24}$
For $MgC{l}_2$
Given, number of moles = 0.5
$\therefore$ Number of particles $=5\times 6.022\times {10}^{23}=3.011\times {10}^{23}$
Mass = number of moles $\times$ molar mass
(molar mass of $MgC{l}_2=$ 24+2 $\times$ 25.5=24+71=95)
$=0.5\times 95=47.5g$
Thus, the completed tabe in as
$\begin{array}{ccccc}Speciesproperty& H_{2}O& C{O}_2& Na-atom& MgC{l}_2\\ Numberofmoles& 2& 0.5& 5.0& 0.5\\ Numberofparticles& 1.2011\times {10}^{24}& 3.011\times {10}^{23}& 3.011\times {10}^{24}& 3.011\times {10}^{23}\\ Mass& 36g& 22g& 115g& 47.5g\end{array}$