Join WhatsApp Icon JEE WhatsApp Group
Question 53

4 g equimolar mixture of NaOH and Na$$_2$$CO$$_3$$ contains xg of NaOH and yg of Na$$_2$$CO$$_3$$. The value of x is ___g. (Nearest integer)


Correct Answer: 1

In an equimolar mixture, both NaOH and Na$$_2$$CO$$_3$$ are present in equal numbers of moles. Let the number of moles of each compound be $$n$$. The total mass of the mixture is 4 g:

$$n \times M_{\text{NaOH}} + n \times M_{\text{Na}_2\text{CO}_3} = 4 \text{ g}$$

The molar masses are: NaOH = 23 + 16 + 1 = 40 g mol$$^{-1}$$, and Na$$_2$$CO$$_3$$ = 2(23) + 12 + 3(16) = 106 g mol$$^{-1}$$.

$$n(40 + 106) = 4$$

$$n = \frac{4}{146} = 0.02740 \text{ mol}$$

The mass of NaOH in the mixture is:

$$x = n \times 40 = \frac{4}{146} \times 40 = \frac{160}{146} \approx 1.096 \text{ g}$$

Rounding to the nearest integer, the value of $$x$$ is $$\mathbf{1}$$ g.

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.