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1 Answer
Apr 2, 2018

#"0.160 M"#

Explanation:

Molar mass of #"B"_2"Br"_6# is #"501.05 g/mol"#

Moles of #"16 g"# of #"B"_2"Br"_6# is #= "16 g"/"501.0 g/mol" = "0.03194 mol"#

#underbrace("B"_2"Br"_6) + "6HNO"_3 -> underbrace("2B(NO"_3")"_3)+ "6HBr"#
#color(blue)"1 mol" color(white)(.........................)color(blue)"2 mol"#

From above equation, #"1 mol"# of #"B"_2"Br"_6# on reaction with nitric acid forms #"2 mol"# of #"B(NO"_3")"_3#.

It’s given that #"16 g"\ ("0.03194 mol")# of #"B"_2"Br"_6# is reacted

So, moles of #"B(NO"_3")"_3# formed #= 2 × "0.03194 mol" = "0.06388 mol"#

#"Molarity" = "Moles of solute"/"Volume of solution (in litres)"#

#"Molarity" = "0.06388 mol"/"0.4 L" = "0.1597 M" ≈ "0.160 M"#