Basic Concepts of Chemistry — NEET UG practice

1. Recall the formula of urea and how many hydrogen atoms it contains.

Urea is $\mathrm{NH_2CONH_2}$, also written as $\mathrm{CO(NH_2)_2}$. Counting the hydrogens in this formula, each molecule of urea contains 4 hydrogen atoms.

2. Convert the given mass of urea into moles.

To count atoms, we first need the number of moles of urea, using:

$$\text{moles} = \frac{\text{given mass}}{\text{molar mass}}$$

Given mass is $5.4\ \mathrm{g}$ and molar mass is $60\ \mathrm{g\ mol^{-1}}$, so:

$$\text{moles of urea} = \frac{5.4}{60} = 0.09\ \mathrm{mol}$$

3. Convert moles of urea into moles of hydrogen atoms.

Since each mole of urea contains $4$ moles of hydrogen atoms, we multiply:

$$\text{moles of H atoms} = 0.09 \times 4 = 0.36\ \mathrm{mol}$$

4. Convert moles of hydrogen atoms into actual number of atoms using Avogadro's number.

Avogadro's number tells us how many particles are present in one mole, so:

$$\text{Number of H atoms} = 0.36 \times 6.022 \times 10^{23} \approx 2.168 \times 10^{23}$$

Hence, the number of hydrogen atoms present is $2.168 \times 10^{23}$, matching Option D.

1. Write the reaction between $\mathrm{CO_2}$ and hot coke (carbon).

When carbon dioxide is passed over red-hot coke, it partly reacts to form carbon monoxide:

$$\mathrm{CO_2} + \mathrm{C} \rightarrow 2\mathrm{CO}$$

2. Recall why volumes can represent moles here.

Since temperature and pressure are the same (STP) before and after the reaction, Avogadro's law tells us that volume of a gas is directly proportional to its number of moles. So we can work directly with volumes instead of converting to moles.

3. Let a variable represent the unknown volume of $\mathrm{CO_2}$ that reacts.

Let $𝑥\ \mathrm{dm^3}$ of $\mathrm{CO_2}$ be consumed in the reaction. From the balanced equation, every $1$ volume of $\mathrm{CO_2}$ that reacts produces $2$ volumes of $\mathrm{CO}$, so:

$$\text{CO}_2 \text{ consumed} = 𝑥, \qquad \text{CO formed} = 2𝑥$$

4. Set up the total volume after reaction.

Initially there was $1\ \mathrm{dm^3}$ of $\mathrm{CO_2}$. After the reaction, unreacted $\mathrm{CO_2}$ is $(1-𝑥)$ and the newly formed $\mathrm{CO}$ is $2𝑥$. Since the final total volume is given as $1.4\ \mathrm{dm^3}$:

$$(1-𝑥) + 2𝑥 = 1.4$$

5. Solve for $𝑥$.

Simplifying the equation:

$$1 + 𝑥 = 1.4 \implies 𝑥 = 0.4$$

6. Find the final volumes of $\mathrm{CO}$ and $\mathrm{CO_2}$.

Therefore, using $𝑥 = 0.4$:

$$\text{Volume of CO} = 2𝑥 = 0.8\ \mathrm{dm^3}, \qquad \text{Volume of } \mathrm{CO_2} = 1 - 𝑥 = 0.6\ \mathrm{dm^3}$$

Hence, the gaseous mixture contains $0.8\ \mathrm{dm^3}$ of $\mathrm{CO}$ and $0.6\ \mathrm{dm^3}$ of $\mathrm{CO_2}$, so the answer is B.

1. Recall the rounding-off rule for a digit exactly equal to 5.

When the digit that has to be dropped is exactly $5$ (with no other non-zero digit after it), chemists use the "round half to even" rule instead of always rounding up. This rule says:

$$\text{if the preceding digit is even, keep it unchanged; if it is odd, increase it by 1}$$

This rule avoids a systematic upward bias that would occur if every $5$ were simply rounded up.

2. Apply the rule to $17.0145$.

Here the digit to be removed is the last $5$, and the digit just before it is $4$, which is even. So, by the rule, the $4$ stays as it is.

$$17.0145 \rightarrow 17.014$$

3. Apply the rule to $21.0235$.

Here again the digit to be removed is the last $5$, and the digit just before it is $3$, which is odd. So, by the rule, the $3$ must be increased by 1.

$$21.0235 \rightarrow 21.024$$

4. Match with the options.

The rounded values are $17.014$ and $21.024$, which matches Option D.

Hence, the answer is D.