Question

Please answer these 2 questionsZz !!

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Answer 1

ANSWER–1

Option A is correct.

Explanation:

We know that volume of

$\begin{gathered}\sf {10}^{3} kg \: of \: water \: = 1 {m}^{3} \\ \sf= > volume \: of \: 1kg \: water = \frac{1}{1000} {m}^{3} = (\frac{1}{ {10}} )^{3} \\\sf or \: = \: \sf one \: decimeter \: cube\end{gathered}$

• And we also know that 1dm^3 = one litre

$\sf Hence \; the \; mass \; of\; one\; kg \; water \; = 1 litre$

And one gram = 1 mL

We also know that one mole of water = 18 gram = 18mL contain 6.0 x 1023 molecules

• Hence the volume of one molecule of water

$\sf= \frac{18}{6 \times {10}^{23} } \: ml = 3 \times {10}^{ - 23}$

_______________________

Molar Mass of the Al₂SO₄.18H₂O is 474 g/mole

474 grams of Aluminium Sulphate Crystals Contains 324 grams of Water.

1 grams--→324/474 grams of the Water.

333 grams --→ 227.62 grams of Water.

No. of Molecules of Water = (Mass/Molecular mass) × Nₐ

• = (227.62/18) × 6.022 × 10²³.
• = 76.15 × 10²³ Molecules.

Thus, the No. of Molecules of Water is 7.15 × 10²³.

Answer 2

$\large { \underline {\underline {\sf ( \:3\:)\:}}}$

Given that , The Density of water at 4⁰ C is 1.0 × 10³ kg/m³ .

Need To Find : The volume occupied by one molecule of water is ______ ?

As , We know that ,

⠀⠀⠀⠀⠀▪︎⠀Molar mass of Water is 18 gm/mole .

$\qquad \therefore \sf Number _{( \:Moles \: in \: 1 \:gm \:)} \:=\:\dfrac{1}{18}\:\\\\ \qquad :\implies \sf Number _{( \:Moles \: in \: 1 \:gm \:)} \:=\:\dfrac{1}{18}\:\\\\\qquad :\implies \sf Number _{( \:Moles \: in \: 1 \:gm \:)} \:=\:0.55 \:$

⠀⠀⠀AND ,

⠀⠀⠀⠀⠀▪︎⠀ 18 gm contains 6.022 × 10²³ molecules of H₂O .

$\qquad \therefore \sf Molecules \:of \:H_2O \: _{( \:\: 1 \:gm \:)} \:=\: \dfrac{6.022 \times 10 ^{23} }{18}\:\\\\ \qquad :\implies \sf Molecules \:of \:H_2O \: _{( \:\: 1 \:gm \:)} \: \:=\:\: \dfrac{6.022 \times 10 ^{23} }{18} \:\\\\\qquad :\implies \sf Molecules \:of \:H_2O \: _{( \:\: 1 \:gm \:)} \: \:=\:0.33 \times 10^{23}\:$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Volume occupied by 1 molecule of water :

$\qquad \dashrightarrow \sf Volume \:Occupied \:_{(\: 1\;molecule \:of \:water \:)}\:=\:\dfrac{1}{0.33 \times 10 ^{23} }\\\\\qquad \dashrightarrow \sf Volume \:Occupied \:_{(\: 1\;molecule \:of \:water \:)}\:=\:\dfrac{1}{0.33 \times 10 ^{23} }\\\\\qquad \dashrightarrow \sf Volume \:Occupied \:_{(\: 1\;molecule \:of \:water \:)}\:=\:2.99 \: \times 10^{-23} \:\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\sf{\sf Volume \:Occupied \:_{(\: 1\;molecule \:of \:water \:)}\:\approx\:3.0 \: \times 10^{-23} }}}}\:\:\bigstar\:$

$\qquad \therefore \underline {\sf Hence, \:Volume \:occupied \:by \:molecule \:of \:water \:is \:\pmb{\bf Option \: A \:)\: 3.0 \: \times 10^{-23} \: }\:approximately \:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\large { \underline {\underline {\sf ( \:4\:)\:}}}$

Given that , The mass of Al₂(SO₄)₃ .18H₂O is 333 gm .

Need To Find : The number of molecules of water in 333 gm of Al₂(SO₄)₃ .18H₂O is ______ ?

As , we know that ,

$\qquad \therefore \sf Molar \:Mass \:_{(\:Al_2(SO_4)_3 .18 H_2O \:)} = ( 2 \times 27 ) + 3 \times \{ 32 + ( 3 \times 17 )\} + ( 1 8 \times 18 ) \\\\ \qquad \dashrightarrow \sf Molar \:Mass \:_{(\:Al_2(SO_4)_3 .18 H_2O \:)} = ( 2 \times 27 ) + 3 \times \{ 32 + ( 3 \times 17 )\} + ( 1 8 \times 18 ) \\\\ \qquad \dashrightarrow \sf Molar \:Mass \:_{(\:Al_2(SO_4)_3 .18 H_2O \:)} = 666 \:g/mol \:$

So,

$\qquad \therefore \sf \:No. \:of \:Moles \:=\: \dfrac{ Given \:mass }{Molar \:mass \:} \\\\ \qquad \dashrightarrow \sf \:No. \:of \:Moles \:=\: \dfrac{ Given \:mass }{Molar \:mass \:}\\\\ \qquad \dashrightarrow \sf \:No. \:of \:Moles \:=\: \dfrac{ 333 }{666\:}\\\\ \qquad \dashrightarrow \sf \:No. \:of \:Moles \:=\: 0.5 \:moles \:$

Then ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding number of molecules of water in 333 gm of Al₂(SO₄)₃ .18H₂O :

$\qquad :\implies \sf Molar \:Mass_{(Water) } \times No. \:of \:moles \: \times \:Avogadro's \:number \:\\\\ \qquad :\implies \sf 18 \times 0.5 \: \times \:6.02 \times 10^{23} \:\\\\ \qquad :\implies \sf 9\: \times \:6.02 \times 10^{23} \:\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\sf{\sf Number \:of \:molecules \: \:=\:9\: \times \:6.02 \times 10^{23} }}}}\:\:\bigstar\:$

$\qquad \therefore \underline {\sf Hence, \:number \: of\: \:molecules \:of \:water \:in \:\:Al_2(SO_4)_3 .18 H_2O \: \pmb{\bf Option \: B \:)\: \:9\: \times \:6.02 \times 10^{23} \: }\: \:.}$