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Calculate the mass of \mathrm{\mathrm{NaCl}} and \mathrm{\mathrm{NaBr} } present in a \mathrm{15 \mathrm{~g}} mixture if out of the total weight of the mixture \mathrm{30 \%} of sodium is present. (Given atomic weight of \mathrm{\mathrm{Na}, \mathrm{Cl} \ \&\ \mathrm{Br}}as \mathrm{23, 35.5 \times 80}, respectively).

Option: 1

\mathrm{\mathrm{NaCl}=7.22 \mathrm{~g}, \mathrm{NaBr}=7.78 \mathrm{~g}}


Option: 2

\mathrm{\mathrm{NaCl}=8.22 \mathrm{~g}, \mathrm{NaBr}=6.77 \mathrm{~g}}


Option: 3

\mathrm{\mathrm{NaCl}=7 \mathrm{~g}, \mathrm{NaBr}=8 \mathrm{~g}}


Option: 4

\mathrm{\mathrm{NaCl}=6.77 \mathrm{~g}, \mathrm{NaBr}=8.22 \mathrm{~g}}


Answers (1)

best_answer

Let NaCl present m the mixture be  x g

\mathrm{\therefore \quad \text { NaBr present }=(15-x) \mathrm{g}}

\mathrm{\text { Formula weight of } \mathrm{NaCl}=58.5}

\mathrm{\text { Formule weight of } \mathrm{NaBr}=103 \mathrm{~g}}

\mathrm{\therefore \text { Total amount of Na present }=\frac{23}{58.5} x+\frac{23}{103}(15-x)}

                                                                      =0.30 \times 15=4.5

\mathrm{\frac{x}{58.5}+\frac{15-x}{103}=\frac{4.5}{23}}

\mathrm{\frac{103 x+877.5-58.5 x}{58.5 \times 103}=\frac{4.5}{23}}

\mathrm{44.5 x+877.5=\frac{4.5 \times 58.5 \times 103}{28}}

\mathrm{x=6.77 \mathrm{~g}=\text { Amount of } \mathrm{NaCl}}

\mathrm{15-x=8.22 \mathrm{~g}=\text { Amount of } \mathrm{NaBr}}

 

Posted by

himanshu.meshram

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