Question

There are 10 g of mixture of NaCl and NaBr. If the amount
of sodium is 25% of the weight of total mixture, calculate
the amount of NaCl and NaBr present in the mixture.
(Given, atomic weights of Na, Cl and Br are 23, 35.5 and
80, respectively).
(1) NaCl = 1.5734 g, NaBr = 1.5734 g
(2) NaCl = 8.4266 g, NaBr = 1.5734 g
(3) NaCl = 8.4266 g, NaBr = 8.4266 g
(4) NaCl = 1.5734 g, NaBr = 8.4266 g​

Answer 1

Answer:

NaCl=1.5734g,NaBr=8.42266g

Answer 2

Answer:

The amount of $NaCl$ in the mixture, $w_{NaCl}$ = $xg$ = $1.5718g$.

The amount of $NaBr$ in the mixture, $w_{NaBr}$ = $(10-x)g$ = $(10-1.5718)g$ = $8.482g$.

Therefore, option 4) $NaCl$ = $1.5734 g$, $NaBr$ = $8.4266g$ is correct. ( closest to $1.5718g$ and $8.482g$ ).

Explanation:

Given data,

The total weight of the mixture ( $NaCl$ and $NaBr$ ), W = $10g$

The amount of sodium in the mixture, $w_{Na}$ = $25\%$ of $10g$ = $\frac{25}{100}\times 10$ = $2.5g$

The atomic weight of Na = $23u$

The atomic weight of Cl = $35.5u$

The atomic weight of Br = $80u$

The amount of $NaCl$ in the mixture, $w_{NaCl}$ =?

The amount of $NaBr$ in the mixture, $w_{NaBr}$ =?

Now,

• Let the amount of $NaCl$ in the mixture, $w_{NaCl}$ = $xg$

Thus,

• The amount of $NaBr$ in the mixture, $w_{NaBr}$ = $(10-x)g$

Now,

• The molar mass of $NaCl$ = atomic mass of Na + atomic mass of Cl = $23+35.5$ = $58.5g/mol$

Therefore,

• The number of moles of $NaCl$ = $\frac{Mass}{Molar mass}$ = $\frac{x}{58.5} mol$

And,

• Weight of $Na$ in $NaCl$ = number of moles of $NaCl$ × atomic weight of Na = $\frac{x}{58.5} \times 23$  -------equation (1)

Similarly,

• The molar mass of $NaBr$ = atomic mass of Na + atomic mass of Br = $23+80$ = $103g/mol$

Therefore,

• The number of moles of $NaBr$ = $\frac{Mass}{Molar mass}$ = $\frac{10-x}{103} mol$

And,

• Weight of $Na$ in $NaBr$ = number of moles of $NaBr$ × atomic weight of Na = $\frac{10-x}{103} \times 23$  -------equation (2)

As given,

• Weight of $Na$ in $NaCl$ + Weight of $Na$ in $NaBr$ =  amount of sodium in the mixture
• $\frac{x}{58.5} \times 23$ + $\frac{10-x}{103} \times 23$ = $2.5$
• $\frac{23x}{58.5} + \frac{230-23x}{103} =2.5$
• $\frac{2369x+ 13455- 1345.5x}{6025.5} = 2.5$
• $1023.5x+ 13455 = 2.5\times 6025.5$
• $1023.5x+ 13455 = 15063.75$
• $1023.5x =1608.75$
• x = $1.5718$

Hence,

• The amount of $NaCl$ in the mixture, $w_{NaCl}$ = $xg$ = $1.5718g$.

• The amount of $NaBr$ in the mixture, $w_{NaBr}$ = $(10-x)g$ = $(10-1.5718)g$ = $8.482g$.