Get Answers to all your Questions

header-bg qa

A solid having density of 9\times 10^{3} kg m^{-3} forms face centred cubic crystals of edge length 200\sqrt{2} pm . What is the molar mass of the solid ?

[ Avogadro constant \cong 6\times 10^{23} mol ^{-1}, \pi \cong 3]

  • Option 1)

     0.0216 kg mol^{-1}

  • Option 2)

    0.0305\: kg\: mol^{-1}

  • Option 3)

    0.0432kg \: \: mol^{-1}

  • Option 4)

     0.4320\: kg \: mol^{-1}

Answers (1)

best_answer

 

Density of cubic unit cell -

d = \frac{zM}{a^3N_o}

 

- wherein

Where,

d = density of crystal

z = no. of effective constituent particles in one unit cell

M = molecular weight

a = edge length of unit cell

No = 6.022*1023

As we learned

d= \frac{z\times M}{N_{A}\times a^{3}}\, \, \,\left ( Formula \right )

9\times 10^{3}=\frac{4\times \frac{M_{0}}{6\times 10^{23}}}{\left ( 200\times \sqrt{2}\times 10^{-12} \right )^{3}}

9\times 10^{3}= \frac{4\times M_{0}}{2^{3}\times 2\times 6\times 10^{23}\times \sqrt{2}\times 10^{-30}}

M_{0} = 9\times 6\times 10^{-4}\times 4\times \sqrt{2}=0.03Kg/mol

 


Option 1)

 0.0216 kg mol^{-1}

Option 2)

0.0305\: kg\: mol^{-1}

Option 3)

0.0432kg \: \: mol^{-1}

Option 4)

 0.4320\: kg \: mol^{-1}

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE