1 L of N_2 and frac{7}{8} L of O_2 gases ar | vedclass

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(C) At constant temperature and pressure,the volume is directly proportional to the number of moles: $\frac{V_1}{V_2} = \frac{n_1}{n_2}$.Given $V_{N_2

Vedclass · Accepted Answer

$m_{N_2} = m_{O_2}$

Vedclass · Answer

(C) At constant temperature and pressure,the volume is directly proportional to the number of moles: $\frac{V_1}{V_2} = \frac{n_1}{n_2}$.Given $V_{N_2} = 1 \ L$ and $V_{O_2} = \frac{7}{8} \ L$.Since $n = \frac{m}{M}$,where $m$ is mass and $M$ is molar mass:$\frac{V_{N_2}}{V_{O_2}} = \frac{n_{N_2}}{n_{O_2}} = \frac{m_{N_2} / M_{N_2}}{m_{O_2} / M_{O_2}}$$\frac{1}{7/8} = \frac{m_{N_2} / 28}{m_{O_2} / 32}$$\frac{8}{7} = \frac{m_{N_2}}{m_{O_2}} 	imes \frac{32}{28}$$\frac{8}{7} = \frac{m_{N_2}}{m_{O_2}} 	imes \frac{8}{7}$Therefore,$m_{N_2} = m_{O_2}$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

$1 \ L$ of $N_2$ and $\frac{7}{8} \ L$ of $O_2$ gases are mixed at the same temperature and pressure. What is the relationship between their masses in the mixture?

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