
How many litres of liquid $CC{{l}_{4}}$(d=1.5g/cc) must be measured out to contain $1\times 1{{0}^{25}}$ $CC{{l}_{4}}$ molecules?
(A) 2.35L
(B) 1.89L
(C) 1.7L
(D) None of these.
Answer
553.8k+ views
Hint: Using the formula of the density as, $density=$ $\dfrac{mass}{volume}$, we can from here calculate the volume of the liquid in litres which contains the Avogadro number of particles, from this we can calculate the volume occupied by the 1 molecules and then, the volume occupied by the $1\times 1{{0}^{25}}$ molecules of $CC{{l}_{4}}$. Now, solve it.
Complete step by step solution:
We have been given a density of $CC{{l}_{4}}$, after finding its molecular mass, we can easily calculate its volume by applying the density formula.
As we know that , $density=$ $\dfrac{mass}{volume}$
Then if density =1.5 g/cc (given)
Mass of $CC{{l}_{4}}$ $= 12+ 35.5(4)$
$=12 +142$
$=154 g$
Put all these values in equating(1), we get
$1.5 =$ $\dfrac{154}{volume}$
$Volume =$ $\dfrac{154}{1.5}$
$= 102.66L$
Now, we know that $102.66L$ of the liquid contains $1$ mole of molecules of $CC{{l}_{4}}$ i.e. the Avogadro no of particles which is =$6.022\times 1{{0}^{23}}$ molecules, then;
$6.022\times 1{{0}^{23}}$molecules of $CC{{l}_{4}}$ are present in the = 102.66 litres of the liquid
Then ,
I molecule of $CC{{l}_{4}}$ will be present in =$\dfrac{102.66}{6.022\times {{10}^{23}}}$ litres of the liquid
Similarly,
$1\times 1{{0}^{25}}$ molecules of $CC{{l}_{4}}$ will be present in=$\dfrac{102.66}{6.022\times {{10}^{23}}}\times 1\times 1{{0}^{25}}$ litres of the liquid
= $\dfrac{102.66}{6.022}\times 1\times 1{{0}^{25}}\times {{10}^{-23}}$litres of the liquid
=$\dfrac{102.66}{6.022}\times 1{{0}^{2}}$litres of the liquid
=$1.7$ litres of the liquid
So, thus $1.7L$ of the liquid will contain $1\times 1{{0}^{25}}$ molecules of $CC{{l}_{4}}$ in it.
Hence, option (C) is correct.
Note: Avogadro’s law states that the one of any gas occupies the volume of $22.4L$ and contains the Avogadro number of particles i.e. about $6.022\times 1{{0}^{23}}$ molecules at the standard conditions of temperature($25$$^{\circ }C$) and pressure($1 atm$).
Complete step by step solution:
We have been given a density of $CC{{l}_{4}}$, after finding its molecular mass, we can easily calculate its volume by applying the density formula.
As we know that , $density=$ $\dfrac{mass}{volume}$
Then if density =1.5 g/cc (given)
Mass of $CC{{l}_{4}}$ $= 12+ 35.5(4)$
$=12 +142$
$=154 g$
Put all these values in equating(1), we get
$1.5 =$ $\dfrac{154}{volume}$
$Volume =$ $\dfrac{154}{1.5}$
$= 102.66L$
Now, we know that $102.66L$ of the liquid contains $1$ mole of molecules of $CC{{l}_{4}}$ i.e. the Avogadro no of particles which is =$6.022\times 1{{0}^{23}}$ molecules, then;
$6.022\times 1{{0}^{23}}$molecules of $CC{{l}_{4}}$ are present in the = 102.66 litres of the liquid
Then ,
I molecule of $CC{{l}_{4}}$ will be present in =$\dfrac{102.66}{6.022\times {{10}^{23}}}$ litres of the liquid
Similarly,
$1\times 1{{0}^{25}}$ molecules of $CC{{l}_{4}}$ will be present in=$\dfrac{102.66}{6.022\times {{10}^{23}}}\times 1\times 1{{0}^{25}}$ litres of the liquid
= $\dfrac{102.66}{6.022}\times 1\times 1{{0}^{25}}\times {{10}^{-23}}$litres of the liquid
=$\dfrac{102.66}{6.022}\times 1{{0}^{2}}$litres of the liquid
=$1.7$ litres of the liquid
So, thus $1.7L$ of the liquid will contain $1\times 1{{0}^{25}}$ molecules of $CC{{l}_{4}}$ in it.
Hence, option (C) is correct.
Note: Avogadro’s law states that the one of any gas occupies the volume of $22.4L$ and contains the Avogadro number of particles i.e. about $6.022\times 1{{0}^{23}}$ molecules at the standard conditions of temperature($25$$^{\circ }C$) and pressure($1 atm$).
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

How many 5 digit telephone numbers can be constructed class 11 maths CBSE

Draw a well labelled diagram of reflex arc and explain class 11 biology CBSE

What is the difference between noise and music Can class 11 physics CBSE

A current I is flowing through wire PQR This wire is class 11 physics CBSE

Trending doubts
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Swaraj means AComplete Independence B Self rule C Self class 11 social science CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

