In which case is the number of molecules of w | vedclass

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(A) To find the number of molecules,we calculate the number of moles $(n)$ in each case,as $Number \; of \; molecules = n 	imes N_A$.$A$) $18 \; mL$

Vedclass · Accepted Answer

$18 \; mL$ of water

Vedclass · Answer

(A) To find the number of molecules,we calculate the number of moles $(n)$ in each case,as $Number \; of \; molecules = n 	imes N_A$.$A$) $18 \; mL$ of water: Since density $d = 1 \; g/mL$,mass $W = 18 \; g$. Moles $n = \frac{18}{18} = 1 \; mol$. Molecules $= 1 	imes N_A$.$B$) $0.18 \; g$ of water: Moles $n = \frac{0.18}{18} = 0.01 \; mol$. Molecules $= 0.01 	imes N_A$.$C$) $0.00224 \; L$ of water vapour at $STP$: Moles $n = \frac{0.00224}{22.4} = 0.0001 \; mol$. Molecules $= 0.0001 	imes N_A$.$D$) $10^{-3} \; mol$ of water: Moles $n = 0.001 \; mol$. Molecules $= 0.001 	imes N_A$.Comparing the values,$1 	imes N_A$ is the maximum. Thus,the correct option is $A$.

In which case is the number of molecules of water maximum?

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