| (i) \[3.0115 \times 1{{0}^{23}}\] molecules of white phosphorus |
| (ii) 10 moles of \[{{H}_{2}}\] gas |
| (iii) 1 g molecule of anhydrous \[N{{a}_{2}}C{{O}_{3}}\] |
| (iv) 33.6 L of \[C{{O}_{2}}\] gas at S.T.P. |
A) \[\left( ii \right)<\left( i \right)<\left( iv \right)<\left( iii \right)\]
B) \[\left( iii \right)>\left( iv \right)>\left( i \right)>\left( ii \right)\]
C) \[\left( i \right)<\left( ii \right)<\left( iv \right)<\left( iii \right)\]
D) \[\left( i \right)<\left( iv \right)<\left( iii \right)<\left( ii \right)\]
Correct Answer: A
Solution :
At. wt. of \[P=31\] and atomicity of P in while P is 4 Mol. wt. of white \[\operatorname{P}= 31 \times 4 = 124\] \[\therefore \,\,\, 6.023 \times 1{{0}^{23}}\] molecules of white P weight 124 g \[\therefore \,\,\,\,\,\,3.0115 \times 1{{0}^{23}}\] molecules of white P weight 124 g \[\frac{124}{6.023}\times 3.0115\,\,=\,\,62\,g\,\,\] (ii) Wt. of mole of \[{{\operatorname{H}}_{2}} gas = 2\,g\] Wt. of 10 moles of \[{{\operatorname{H}}_{2}} gas = 2 \times 10 = 20 g\] (iii) 1 g molecule of anhydrous \[N{{a}_{2}}C{{O}_{3}}=Mol\,wt.\] of \[{{\operatorname{Na}}_{2}}C{{O}_{3}} in\,\,g = 106 g\] (iv) At STP 22.4 L \[C{{O}_{2}}\], weighs 44 g \[\therefore \,\,\, 33.6 L\,C{{O}_{2}}, weighs \frac{44}{22.4} \times 33.6 g = 66 g\] So, the correct choice \[\operatorname{ii} < i < iv < iii\]
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