A volume of \(10 \mathrm{ml}\) of an oxide of nitrogen was taken in a eudiometer tube and mixed with hydrogen until the volume was \(28 \mathrm{ml} .\) On sparking, the resulting mixture occupied \(18 \mathrm{ml}\). To this mixture, oxygen was added when the volume came to \(27 \mathrm{ml}\) and on explosion again, the volume fall to \(15 \mathrm{ml}\). Find the molecular weight of the oxide of nitrogen originally taken in eudiometer tube. All measurements were made at STP.

Let's analyze each part of the given exercise. 1. A volume of 10 ml of an unknown oxide of nitrogen is mixed with hydrogen until the total volume is 28 ml. 2. After sparking, the volume of the mixture is reduced to 18 ml. 3. Oxygen is added to the mixture, increasing the total volume to 27 ml. 4. After another explosion, the volume is reduced to 15 ml.

We are given that the initial volume of the oxide of nitrogen was 10 ml, and hydrogen gas was mixed until the total volume reached 28 ml. Therefore, the volume of hydrogen gas mixed initially can be calculated by subtracting the volume of oxide of nitrogen from the total volume. Volume of hydrogen gas = Total volume - Volume of oxide of nitrogen Volume of hydrogen gas = 28 ml - 10 ml = 18 ml

After the sparking, the volume reduced to 18 ml, and then oxygen was added to the mixture which increased the total volume to 27 ml. The volume of oxygen gas mixed can be calculated as follows: Volume of oxygen gas = Volume after adding oxygen - Volume after sparking Volume of oxygen gas = 27 ml - 18 ml = 9 ml

After the second explosion, the total volume fell to 15 ml. As there is no hydrogen left in the mixture, we can conclude that there are 5 ml nitrogen gas (since 15 ml - 10 ml = 5 ml). Also, we have 9 ml of oxygen gas mixed, and the remaining volume after explosion is 15 ml, which implies that 6 ml of oxygen gas reacted (since 9 ml - (15 ml - 10 ml) = 6 ml). From the data, we can deduce the following information: - From the first reaction, 5 ml of nitrogen gas is produced (total volume fall from the second reaction). This means that 5 ml of the unknown oxide of nitrogen gas reacted with hydrogen to form nitrogen and water. - From the second reaction, 6 ml of oxygen reacted with the remaining 5 ml of the unknown oxide of nitrogen to form nitrogen dioxide.

Let's assume that the formula of the unknown oxide of nitrogen is NxOy. Since 5 ml of NxOy reacts with hydrogen, we have: 5 ml of NxOy -> 5 ml of N2 5 ml of NxOy -> (y * 5) ml of H2O Since the hydrogen volume is 18 ml, y * 5 ml = 18 ml y = 3.6 In the second reaction, when oxygen reacts with the remaining 5 ml of NxOy, the equation is as follows: 5 ml of NxOy + 6 ml of O2 -> 5 ml of N2O5 We can then deduce that: x = 2 y = 3.6 Therefore, the formula of the unknown oxide of nitrogen is N2O3.6 (rounded to one decimal place). We can then calculate the molecular weight as follows: Molecular weight of N2O3.6 = (2 * Molecular weight of N) + (3.6 * Molecular weight of O) Molecular weight of N2O3.6 = (2 * 14) + (3.6 * 16) Molecular weight of N2O3.6 = 28 + 57.6 Molecular weight of N2O3.6 = 85.6 g/mol