Question

calculate the mass of calcium that will contain the same number of atoms as are present in 3.2g of sulphur(atomic masses:S=32,Ca=40)

Answer 1

first, let us calculate the number of atoms present in the given sample of sulphur.
$no. of mol = \frac{given mass}{molar mass}$
∴$n= \frac{3.2}{32} \\ n= 0.1$
$No. of atoms= n *Avogadro's. constant \\ atoms= 0.1*6.022* 10^{23} \\ atoms=0.6022* 10^{23} \\ atoms= 6.0228*10^{24}$

Now we know the no. of atoms in the given sample of Sulphur.
to find the mol of Calcium with equal atoms,
we need to divide the obtained result by Avogadro's constant

$\frac{6.022* 10^{24} }{6.022* 10^{23} } = 0.1$

now, we know the mols
$n= \frac{given mass}{molar mass} \\ 0.1 = \frac{given mass}{40} \\ 0.1*40= given mass \\ 4g = given mass$

∴ 4g of Calcium will have the same number of atoms as 3.2g of Sulphur.

Answer 2

it may not be neat....

it is the most prominent and shortest CORRECT ANSWER to this question...