Question

') 2.0 2. 10. In which case is number of molecules of water maximum? [NEET-2018] (1) 18 mL of water (2) 0.18 g of water (3) 10-3 mol of water (4) 0.00224 L of water vapours at 1 atm and 273K ombine to form

NEET/Medical Exams

Chemistry

Solution

149

4.0 (1 ratings)

(1) ( 18 mathrm{mL} ) water ( begin{aligned} text { As } mathrm{d}_{mathrm{H}_{2} mathrm{O}}=1 mathrm{g} / mathrm{mL} & text { So } mathrm{W}_{mathrm{H}_{2} mathrm{O}}=18 mathrm{g} mathrm{n}_{mathrm{H}_{2} mathrm{O}} &=frac{18}{18}=1 & text { molecules }=1 times mathrm{N}_{mathrm{A}} end{aligned} ) (2) 0.18 g of water ( mathrm{n}_{mathrm{H}_{2} mathrm{O}}=frac{0.18}{18}=0.01 ) (molecules) ( mathrm{H}_{2} mathrm{O}=0.01 times mathrm{N}_{mathrm{A}} ) ( (3)left(mathrm{V}_{mathrm{H}_{2} mathrm{O}(mathrm{g})}right)_{mathrm{STP}}=0.00224 mathrm{L} ) ( mathrm{n}_{mathrm{H}_{2} mathrm{O}}=frac{mathrm{V}}{22.4}=frac{0.00224}{22.4}=0.0001 ) molecules ( =0.0001 times mathrm{N}_{mathrm{A}} ) (4) ( mathrm{n}_{mathrm{H}_{2} mathrm{O}}=10^{-3} ) (molecules) ( mathrm{H}_{2} mathrm{O}=10^{-3} times mathrm{N}_{mathrm{A}} )