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}"
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