A mixture of CO2 and H2 contains 4-5 - 10 23 molecules in it- The partial pressure of CO2 is 23rd of the total pressure- The weight of CO2 present in the mixture in grams is- - N - 6 - 1023-

The correct option is C 22 P_CO_2/P=X_CO_2=2/3 = moles of CO_2/Total moles of mixture Total moles of mixture = 4.5 × 10^23 6 × 10^23 = 34 #160; ...

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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

A mixture of ...

Question

A mixture of $C O_{2}$ and $H_{2}$ contains 4.5 $\times$ 10 $^{23}$ molecules in it. The partial pressure of $C O_{2}$ is $\frac{2}{3} r d$ of the total pressure. The weight of $C O_{2}$ present in the mixture in grams is:

[ $N$ $=$ $6$ $\times$ $10^{23}$ ]

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Solution

The correct option is C 22
$\frac{P_{C O_{2}}}{P} = X_{C O_{2}} = \frac{2}{3} = \frac{m o l e s o f C O_{2}}{T o t a l m o l e s o f m i x t u r e}$

Total moles of mixture $= \frac{4.5 \times 10^{23}}{6 \times 10^{23}} = \frac{3}{4}$

$X_{C O_{2}} = \frac{2}{3} = \frac{m o l e s o f C O_{2}}{\frac{3}{4}}$

Thus, moles of $C O_{2} = 0.5$

Wt. of $C O_{2} = 22 g m$

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A mixture of CO2 and H2 contains 4.5 × 1023 molecules in it. The partial pressure of CO2 is 23rd of the total pressure. The weight of CO2 present in the mixture in grams is: [N = 6 × 1023]

The correct option is C 22PCO2P=XCO2=23=moles of CO2Total moles of mixtureTotal moles of mixture =4.5×10236×1023=34 XCO2=23=moles of CO234Thus, moles of CO2=0.5Wt. of CO2=22 gm

A mixture of $C O_{2}$ and $H_{2}$ contains 4.5 $\times$ 10 $^{23}$ molecules in it. The partial pressure of $C O_{2}$ is $\frac{2}{3} r d$ of the total pressure. The weight of $C O_{2}$ present in the mixture in grams is:

[ $N$ $=$ $6$ $\times$ $10^{23}$ ]