(B) Given: $W = 4 \ g$,$V = 5.6035 \ L$,$T = 546 \ K$,$P = 2 \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.Using the ideal gas equation: $PV = nRT

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(B) Given: $W = 4 \ g$,$V = 5.6035 \ L$,$T = 546 \ K$,$P = 2 \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.Using the ideal gas equation: $PV = nRT = \frac{W}{M} RT$.Rearranging for molecular weight $M$: $M = \frac{WRT}{PV}$.Substituting the values: $M = \frac{4 \times 0.0821 \times 546}{2 \times 5.6035}$.$M = \frac{179.2644}{11.207} \approx 16 \ g \ mol^{-1}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

$4 \ g$ of an ideal gas occupies $5.6035 \ L$ of volume at $546 \ K$ and $2 \ atm$ pressure. What is its molecular weight?

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A
$4$
B
$16$
C
$32$
D
$64$

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$4 \ g$ of an ideal gas occupies $5.6035 \ L$ of volume at $546 \ K$ and $2 \ atm$ pressure. What is its molecular weight?

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