Exercise 0.043 mol of a gas sample weighs 1.892 g. What is the density (in g/L of this gas at 0.0^circ C and 9.55 psi? (1.0atm=14.7psi)

To find the density of the gas sample, we can use the ideal gas law equation:<br /><br />\[ PV = nRT \]<br /><br />where:<br />- \( P \) is the pressure,<br />- \( V \) is the volume,<br />- \( n \) is the number of moles,<br />- \( R \) is the ideal gas constant,<br />- \( T \) is the temperature in Kelvin.<br /><br />First, we need to convert the given pressure from psi to atm:<br />\[ P = \frac{9.55 \text{ psi}}{14.7 \text{ psi/atm}} = 0.647 \text{ atm} \]<br /><br />Next, we convert the temperature from Celsius to Kelvin:<br />\[ T = 0.0^{\circ}C + 273.15 = 273.15 \text{ K} \]<br /><br />The ideal gas constant \( R \) in units of \( \text{L·atm/(mol·K)} \) is:<br />\[ R = 0.0821 \text{ L·atm/(mol·K)} \]<br /><br />Now we can rearrange the ideal gas law to solve for the volume \( V \):<br />\[ V = \frac{nRT}{P} \]<br /><br />Substitute the known values into the equation:<br />\[ V = \frac{(0.043 \text{ mol}) (0.0821 \text{ L·atm/(mol·K)}) (273.15 \text{ K})}{0.647 \text{ atm}} \]<br /><br />Calculate the volume:<br />\[ V = \frac{(0.043) (0.0821) (273.15)}{0.647} \]<br />\[ V = \frac{0.9737}{0.647} \]<br />\[ V \approx 1.51 \text{ L} \]<br /><br />Finally, we can calculate the density \( d \) of the gas using the formula:<br />\[ d = \frac{\text{mass}}{\text{volume}} \]<br /><br />Given the mass of the gas sample is 1.892 g and the volume is 1.51 L:<br />\[ d = \frac{1.892 \text{ g}}{1.51 \text{ L}} \]<br />\[ d \approx 1.25 \text{ g/L} \]<br /><br />Therefore, the density of the gas at \( 0.0^{\circ}C \) and 9.55 psi is approximately \( 1.25 \text{ g/L} \).