The mass of a mixture containing NaCl and NaB | vedclass

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(C) Let the mass of $NaCl$ be $x \ g$ and the mass of $NaBr$ be $(4.0 - x) \ g$. The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. The molar mass o

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Let the mass of $NaCl$ be $x \ g$ and the mass of $NaBr$ be $(4.0 - x) \ g$. The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. The molar mass of $NaBr = 23 + 80 = 103 \ g/mol$. The mass of $Na$ in $NaCl = x 	imes (23 / 58.5) \ g$. The mass of $Na$ in $NaBr = (4.0 - x) 	imes (23 / 103) \ g$. Total mass of $Na = 30 \% 	ext{ of } 4.0 \ g = 1.2 \ g$. So,$x(23 / 58.5) + (4.0 - x)(23 / 103) = 1.2$. Dividing by $23$: $x / 58.5 + (4.0 - x) / 103 = 1.2 / 23 \approx 0.05217$. $0.01709x + 0.03883 - 0.00971x = 0.05217$. $0.00738x = 0.01334$. $x \approx 1.807 \ g$. Percentage of $NaCl = (1.807 / 4.0) 	imes 100 \approx 45.17 \%$. The closest option is $45 \%$.

The mass of a mixture containing $NaCl$ and $NaBr$ is $4.0 \ g$. If $Na$ is $30 \%$ of the total mixture,the composition of $NaCl$ in the mixture is (in $\%$)

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