Q.34. A mixture of carbon dioxide and oxygen has volume , temperature , pressure and mass . The num

Q: What is the correct answer for this exam Chemistry Q34?

The correct answer is C. Given: $V = 8310 \mathrm{~cm}^3 = 8.31 \times 10^{-3} \mathrm{~m}^3$, $T = 300 \mathrm{~K}$, $P = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$, $R = 8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$. Using Ideal Gas Law: $PV = n_{\text{total}}RT$. $n_{\text{total}} = \frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \frac{100}{300} = 0.333 \mathrm{~mol}$. Let $n_1$ be moles of $\

Q: Q.34. A mixture of carbon dioxide and oxygen has volume , temperature , pressure and mass . The number of moles of carbon dioxide and oxygen gases in the mixture respectively are (Assume both…

The correct answer is C. Given: $V = 8310 \mathrm{~cm}^3 = 8.31 \times 10^{-3} \mathrm{~m}^3$, $T = 300 \mathrm{~K}$, $P = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$, $R = 8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$. Using Ideal Gas Law: $PV = n_{\text{total}}RT$. $n_{\text{total}} = \frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \frac{100}{300} = 0.333 \mathrm{~mol}$. Let $n_1$ be moles of $\

Q: What are the options for this this exam Chemistry question?

A. 15 and 0.18 · B. 25 and 0.08 · C. 21 and 0.12 · D. 13 and 0.20

Question

Q.34. A mixture of carbon dioxide and oxygen has volume , temperature , pressure  and mass . The number of moles of carbon dioxide and oxygen gases in the mixture respectively are
(Assume both carbondioxide and oxygen gases behave like ideal gases) [R = 8.31J/mol.K]

FirstInTest · Accepted Answer

Given: $V = 8310 \mathrm{~cm}^3 = 8.31 	imes 10^{-3} \mathrm{~m}^3$, $T = 300 \mathrm{~K}$, $P = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$, $R = 8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$.
Using Ideal Gas Law: $PV = n_{	ext{total}}RT$.
$n_{	ext{total}} = \frac{PV}{RT} = \frac{10^5 	imes 8.31 	imes 10^{-3}}{8.31 	imes 300} = \frac{100}{300} = 0.333 \mathrm{~mol}$.
Let $n_1$ be moles of $\mathrm{CO}_2$ and $n_2$ be moles of $\mathrm{O}_2$.
$n_1 + n_2 = 0.333$ (Equation 1)
Molar mass of $\mathrm{CO}_2 = 12 + 2 	imes 16 = 44 \mathrm{~g/mol}$.
Molar mass of $\mathrm{O}_2 = 2 	imes 16 = 32 \mathrm{~g/mol}$.
Total mass $m = n_1 M_1 + n_2 M_2 = 44n_1 + 32n_2 = 13.2 \mathrm{~g}$ (Equation 2)
From (1), $n_2 = 0.333 - n_1$.
Substitute into (2): $44n_1 + 32(0.333 - n_1) = 13.2$
$44n_1 + 10.656 - 32n_1 = 13.2$
$12n_1 = 13.2 - 10.656 = 2.544$
$n_1 = \frac{2.544}{12} \approx 0.212 \mathrm{~mol}$.
$n_2 = 0.333 - 0.212 = 0.121 \mathrm{~mol}$.
So, $n_1 \approx 0.21 \mathrm{~mol}$ and $n_2 \approx 0.12 \mathrm{~mol}$. The closest option is (3) 0.21 and 0.12.

Answer

15 and 0.18

Answer

25 and 0.08

Answer

21 and 0.12

Answer

13 and 0.20

Related Questionsin Chemistry

Browse Engineering Questions by Subject

Practice Other Engineering Exams

Practice More Engineering Questions