Calculate the mass of Na_{​​​​​2}​CO_​​3 which will have the same number of molecules as contained in 12.3 gram of MgSO_​​​4​​MgSO4.7H₂O

we know no. of molecules in 12.3 g of MgSo4.7H2O = no. of molecules in some mass of sodium carbonate

so first we need to find the no. of molecules in 12.3 g of MgSo4.7H2O....

molar mass of MgSo4.7H2O = 24 + 32 + 64 + 126 = 246 g / mol

as we know molar mass is nothing but 1 mole only. and 1 mole has 6.022 x 10 raised to the power of 23

therefore 24g/ mol = 6.022 x 10 raised to the power of 23

so 12.3 g will have this no. of molecules = [ 6.022 x 10 raised to the power 23 x 12.3 ] / 246

= 3.011 x 10 raised to the power 22

therefore some mass of sodium carbonate has also got 3.011 x 10 raised to the power of 22 molecules

hence molar mass os sodium carbonate = 46 + 12 + 48 = 106 g / mol

in a similar way

6.022 x 10 raised to the power of 23 molecules are there in = 106g

so, 3.011 x 10 raised to the power of 22 molecules will be in = [ 106 x 3.011 x 10 raised to the power of 22 ] / 6.022 x 10 raised to the power 23

= 5.3 g ..... answer