Question

there is 10g of mixture of NaCl and Na Br. if the amount of sodium is 25% of weight of total mixture calculate amount of Nacl and NaBr present in the mixture?

Answer 1

Question#45195 – Chemistry – Inorganic Chemistry

Question:

There is 10g of mixture of NaCl and NaBr. if the amount of sodium is 25% of weight of total mixture calculate amount of NaCl and NaBr present in the mixture?

Answer:

The total mass of sodium in the mixture:

$$m(Na) = \omega \times 10 = 2.5g$$

Total mass consists of sodium from NaCl and from NaBr.

Let's consider m(NaCl) = x g, than m(NaBr) = 10 - x g.

The weight percent of sodium in NaCl:

$$w_1 = \frac{23}{58.5} = 0.39$$

The weight percent of sodium in NaBr:

$$w_2 = \frac{23}{103} = 0.22$$ $$m_1(Na) = 0.39x$$ m_2(Na) = \frac{0.22(10 - x)}{ $$m(Na) = m_1(Na) + m_2(Na)$$ $$2.5 = 0.39x + 0.22(10 - x)$$ $$2.5 = 0.39x + 2.2 - 0.22x$$ $$0.3 = 0.17x$$ $$x = 1.765$$

The mass of NaCl is 1.76 g. The weight percentage of it in the mixture is:

$$\omega(NaCl) = \frac{1.76}{10} = 0.176 = 17.6\%$$ $$\omega(NaBr) = \frac{100 - 17.6}{100 - 17.6} = 82.4\%$$

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