A volatile liquid (0.146 g) displaced 30.5 cm3 of air at 30.5 cm3 of air at 305 K and 760 mmHg pressure. The molar mass of the liquid is

n=pVRT=(1atm)30.5×10−3L0.0825LatmK−1mol−1(305K)=0.00121molM=mn=0.146g0.00121mol=120.7gmol−1

A volatile liquid (0.146 g) displaced 30.5 cm3 of air at 30.5 cm3 of air at 305 K and 760 mmHg pressure. The molar mass of the liquid is

n=pVRT=(1atm)30.5×10−3L0.0825LatmK−1mol−1(305K)=0.00121molM=mn=0.146g0.00121mol=120.7gmol−1

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A volatile liquid (0.146 g) displaced 30.5 cm³ of air at 30.5 cm³of air at 305 $K$ and 760 $mmHg$ pressure. The molar mass of the liquid is

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$117.8 g {mol}^{- 1}$

$118.8 g {mol}^{- 1}$

$119.8 g {mol}^{- 1}$

$120.8 g {mol}^{- 1}$

answer is D.

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Detailed Solution

$n = \frac{p V}{R T} = \frac{(1 atm) (30.5 \times 10^{- 3} L)}{(0.0825 {LatmK}^{- 1} {mol}^{- 1}) (305 K)} = 0.00121 mol$

$M = \frac{m}{n} = \frac{0.146 g}{0.00121 mol} = 120.7 g {mol}^{- 1}$

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A volatile liquid (0.146 g) displaced 30.5 cm3 of air at 30.5 cm3 of air at 305 K and 760 mmHg pressure. The molar mass of the liquid is

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A volatile liquid (0.146 g) displaced 30.5 cm³ of air at 30.5 cm³of air at 305 $K$ and 760 $mmHg$ pressure. The molar mass of the liquid is