Question

S - 1) The specific weight of the X2Y3 compound is 4.8 g / cm3. The atomic weights of X and Y are 30 and 15 g respectively, the atomic radii are 0.55 and 1.5 Angstroms. Avagadro number: 6.02x1023.


a) How many atoms are there in 1 cm X2Y3?


b) How many atoms are in 1 g of X2Y3.

Answer 1

a)

we solve by the formula

$\rho=\frac{nA}{VcNa}$

p=4.8g / cm3

A=105 g/mol

Na=6.02x1023.  

Vc=a^3

$a=4r\sqrt{2}$

0.55A=0.055 10^-7cm3

1.5A=0.15 10^-7cm3

$a=4\times0.055\times10^{-7}\times\sqrt{2}=0.31\times10^{-7}$

$Vc=(0.31\times10^{-7})^3=0.03\times10^{-21}$

$a=4\times0.15\times10^{-7}\times\sqrt{2}=0.31\times10^{-7}=0.84\times10^{-7}$

$Vc=(0.84\times10^{-7})^3=0.59\times10^{-21}$

$Vc=0.59\times10^{-21}+0.03\times10^{-21}=0.62\times10^{-21}$

$4.8=\frac{n\times105}{0.62\times10^{-21}\times6.02\times10{23}}$

n=17

b)

find the mass

$\gamma=\frac{m}{V}$

$m=\gamma\times V=4.8\times$0.62\times10^{-21}=2.976\times10^{-21}

$n=\frac{m\times M}{Na}=\frac{2.976\times10^{-21}\times105}{6.02\times10^{23}}=51.9\times10^{-44}$