Question

The pressure of sodium vapour in a 1.0 L container is 9.5 torr at 927°C. How many atoms are in the container?
(A) 9.7 x 10^7
(B) 7.5 10^19
(C) 4.2 x 10^17
(D) 9.7 * 10^19

PLS HELP DONT SPAM AND ANSWER WITH FULL EXPLANATION OR WILL BE REPORTED.​

Answer 1

Answer:

Correct option is

B

7.5×10^19

10 torr = 0.0013×10 atm = 0.013 atm

By ideal gas equation,n= PV/RT

= 0.013×1/0.0821×1273

=1.24×10^−4 moles

∴ Number of atoms = 1.24×10^−4×6.022×10^23

= 7.5×10^19 atoms

Hence, option B is correct.

Answer 2

$\boxed{\sf{\purple{Option\:\bf{(B)\:7.5\:\times\:10^{19}}\:\sf{is\:correct!}}}}$

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Explanation :

$\begin{gathered}\begin{gathered}\underline{\bf\dag{\textsf{\textbf{\underline{Given}}}}}\begin{cases} & \sf{Volume_{(sodium\:vapour)} = \frak{1\;l}} \\ \\ & \sf{Temperature = \frak{927^{\circ}C}}\\ \\ & \sf{Pressure = \frak{9.5\;torr}} \end{cases} \end{gathered}\end{gathered}$

$\underline{\bf\dag{\textsf{\textbf{\underline{To\:Find}}}}:-}$

• Number of atoms in the container ?

$\underline{\bf\dag{\textsf{\textbf{\underline{Solution}}}}:-}$

★ Converting units of temperature from celsius to kelvins :-

$\quad\small\bf\dag\:\bigg\lgroup {\sf{\pink{Temperature_{(in\:kelvins)} = (0^{\circ}C + 273)\:K}}}$

$\quad\leadsto\quad\sf Temperature_{(in\:kelvins)} = (927^{\circ}C + 273)\:k$

$\quad\leadsto\quad{\bf{\green {Temperature_{(in\:kelvins)} = 1200\:K}}}$

★ Converting units of pressure from torr to atm :-

$\qquad\bf\dag\:\bigg\lgroup {\sf{\purple{1\:torr = \dfrac{1}{760}\:atm}}}$

$\qquad\leadsto\quad\sf 9.5\:torr = \bigg(9.5\:\times\:\dfrac{1}{760}\bigg)\:atm$

$\qquad\leadsto\quad\sf 9.5\:torr = \dfrac{9.5}{760}\:atm$

$\qquad\leadsto\quad\sf 9.5\:torr = \dfrac{\cancel{9.5}}{\cancel{760}}\:atm$

$\qquad\leadsto\quad{\bf{\red {9.5\:torr = 0.0125\:atm}}}$

★ Using ideal gas equation to find number of moles :-

$\qquad\bf\dag\:\bigg\lgroup {\sf{\orange{PV = nRT}}}$

★ Putting all known values :-

$\quad\longrightarrow\quad\sf 0.0125 \:\times\:1 = n \:\times\: 0.0821\:\times\:1200$

$\quad\longrightarrow\quad\sf 0.0125 = 98.52n$

$\quad\longrightarrow\quad\sf \dfrac{0.0125}{98.52} = n$

$\quad\longrightarrow\quad\sf \dfrac{\cancel{0.0125}}{\cancel{98.52}} = n$

$\quad\longrightarrow\quad{\bf{\blue{n = 0.000126}}}$

Now,

★ Finding number of atoms present in the container :-

$\qquad\tiny\bf\dag\:\bigg\lgroup{\sf{\green {Number\:of\:atoms = Number\:of\:moles\:\times\:Avogadro\:number}}}$

$\small\longrightarrow\:\sf Number\:of\:atoms = 0.000126 \:\times\: 6.022 \:\times\: 10^{23}$

$\small\longrightarrow\:\sf Number\:of\:atoms = 0.00075 \:\times\: 10^{23}$

$\small\longrightarrow\:{\bf{\red{ Number\:of\:atoms = 7.5 \:\times\: 10^{19}}}}$

$\therefore\:{\underline{\sf{Hence,\:number\:of\:atoms\:=\:\bf{7.5 \:\times\: 10^{19}}}}}$

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