
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
581.7k+ 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($1 atm$).
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