Question

What volume of ${CCl}_4(d=1.6g/cc)$ contain $6.02\times {10}^{25}CC{l}_4$ molecules? $(Cl=35.5$)

• A. $10.5L$
• B. $250L$
• C. $9.625L$
• D. $1.712L$

Answer 1

The correct option is C $9.625L$

$1$ mole of $CC{l}_4$ contains $6.022\times {10}^{23}$ molecules

$\therefore$ Moles of $CC{l}_4=\frac{6.02\times {10}^{25}}{6.02\times {10}^{23}}=100$ moles

$\therefore$ Mass of $CC{l}_4=100\times (12+4\times 35.5)=15400g$

$\therefore$ Density=$\frac{Mass}{Volume}$

$\therefore$ Volume=$\frac{15400}{1.6}mL=9.625L$