Which option contains the same number of chlorine atoms as present in a #142*g# mass of #Cl_2#?

Question

#"a. 222 g calcium chloride;"#
#"b. 111 g magnesium chloride;"#
#"c. 444 g calcium chloride;"#
#"d. 234 g sodium chloride?"#

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Answer 1 · 2017-04-28T05:52:25

#"Moles of"# #Cl_2=(142*g)/(70.9*g*mol^-1)=2.00*mol#

Now #NaCl# has a molar mass of #58.44*g*mol^-1#

And thus we need a FOUR MOLE quantity of this salt, i.e. #2*molxx58.44*g*mol^-1=234*g#.

So your option is clear.