Question

$500$ ml of solution of $0.1MNaBr$ contains how many milligrams of bromine?

• A. $200$ mg
• B. $400$ mg
• C. $2000$ mg
• D. $4000$ mg
• E. $20000$ mg

Answer 1

The correct option is D $4000$ mg

$500$ ml of solution corresponds to $\frac{500}{1000}=0.500$ L of solution.

$0.500$ L of solution of $0.1MNaBr$ corresponds to $0.500$ L $\times 0.1$ mol/L $=0.05$ moles of $NaBr$.
$1$ mole of $NaBr$ contains $1$ mole of bromide ion.
$0.05$ moles of $NaBr$ will contain $0.05$ moles of bromide ions. The atomic weight of bromine is 79.9 g/mol. The mass of bromine present is $79.9$ g/mol $\times 0.05$ mol $4$ g
$4$ g of bromine corresponds to $4$ g $\times 1,000$ mg/g $=4000$ mg.
Hence, 500 ml of solution of $0.1MNaBr$ has $4000$ mg of bromine.