Arrange the following in order of increasing mass 

(i) $3.0115\times {10}^{23}$ molecules of white phosphorus

(ii) 10 moles of ${\mathrm{H}}_2 \mathrm{gas}$

(iii) 1 g molecule of anhydrous ${\mathrm{Na}}_2{\mathrm{CO}}_3$

(iv) 33.6 L of ${\mathrm{CO}}_2$ gas at STP

(i) At. wt. of P = 31 and atomicity of P is 4

Mol. wt.of white P = 31 x 4 = 124

$\begin{array}{l}\therefore 6.023\times {10}^{23}\mathrm{molecules}\mathrm{of}\mathrm{white}\mathrm{P}\mathrm{weigh}124\mathrm{g}\\ \therefore 3.0115\times {10}^{23}\mathrm{molecules}\mathrm{of}\mathrm{white}\mathrm{P}\mathrm{weigh}\frac{124}{6.023}\times 3.015=62\mathrm{g}\\ \\ \left(\mathrm{ii}\right)\mathrm{Wt}.\mathrm{of}1\mathrm{mole}\mathrm{of}{\mathrm{H}}_2\mathrm{gas}=2\mathrm{g}\\ \mathrm{Wt}.\mathrm{of}10\mathrm{moles}\mathrm{of}{\mathrm{H}}_2\mathrm{gas}=2\times 10=20\mathrm{g}\\ \left(\mathrm{iii}\right)1\mathrm{g}\mathrm{molecule}\mathrm{of}\mathrm{anhydrous}{\mathrm{Na}}_2{\mathrm{CO}}_3=\mathrm{Mol}\mathrm{wt}.\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3\\ \mathrm{in} \mathrm{g}=106\mathrm{g}\\ \left(\mathrm{iv}\right)\mathrm{At}\mathrm{STP}22.4\mathrm{L}{\mathrm{CO}}_2\mathrm{weighs}44\mathrm{g}\\ \therefore 33.6\mathrm{L}{\mathrm{CO}}_2\mathrm{weighs}\frac{44}{22.4}\times 33.6\mathrm{g}\\ =66\mathrm{g}\\ \mathrm{So},\mathrm{the}\mathrm{correct}\mathrm{choice}\mathrm{ii}<\mathrm{i}<\mathrm{iv}<\mathrm{iii}\end{array}$