Question

How many no. of moles in 52u of helium?

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\boxed{\sf{\purple{1\:mole\:of\:He=6.022\times 10{}^{23}}}}$

$\tt\:A.M.U\:of\: Helium (He)= 4unit\\ \\ \sf\longrightarrow So\:,\:1\:mole\:of\:He=4u\\ \\ \sf\longrightarrow 52u=\cancel{\dfrac{52}{4}}\\ \\ \sf\implies 52u=13moles$

Now the calculate the number of moles

$\sf\implies 1\:mole=6.022\times 10{}^{23}atoms\\ \\ \sf\implies 13\:moles=13\times 6.022\times 10{}^{23}atom\\ \\ \sf\implies 13\:moles=78.286\times 10{}^{23}atom\\ \\ \sf\implies \:\:or\:\: 7.8286\times 10{}^{24}atom$

$\boxed{\sf{\purple{No.\:of\:moles\:in\:52u\:He= 13\:moles}}}$

$\underline{\dag{\sf{\purple{No..\:of\:atom\:=7.8286\times 10{}^{24}atoms}}}}$

Answer 2

we know that ,

1 mole of helium = 6.022×10²³

Atomic mass of helium =4u

=>1 mole of helium =4u

=>52u =52/4

$\mathbf \red{=>52u=13}$

Now,

1 mole =6.022×10²³

=>13 moles=13×6.022×10²³

=>13 moles = 78.286×10²³

$\star \mathbf \red{no. \: of \: moles \: in \: 52u \: of \: he = 78.286 \times 10 {}^{23} }$