Question

a sample of gas has a mass of 0.565 g. its volume is 112 ml at a temperature of 85 °c and a pressure of 763 mm hg. part a find the molar mass of the gas. express your answer in grams per mole to three significant figures. molar mass = g/mol

Explanation:

Step1: Recall Ideal Gas Law

The Ideal Gas Law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant, and \( T \) is temperature (in Kelvin). We can rearrange to find \( n=\frac{PV}{RT} \). Molar mass \( M=\frac{m}{n} \), so \( M = \frac{mRT}{PV} \).

Step2: Convert Units

- Pressure \( P = 763 \, \text{mm Hg} \). Convert to atm: \( 1 \, \text{atm} = 760 \, \text{mm Hg} \), so \( P=\frac{763}{760} \approx 1.004 \, \text{atm} \).
- Volume \( V = 112 \, \text{mL} = 0.112 \, \text{L} \) (since \( 1 \, \text{L} = 1000 \, \text{mL} \)).
- Temperature \( T = 85^\circ \text{C} + 273.15 = 358.15 \, \text{K} \).
- Mass \( m = 0.565 \, \text{g} \).
- Gas constant \( R = 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \).

Step3: Calculate Moles (\( n \))

Using \( n=\frac{PV}{RT} \):
\( n=\frac{1.004 \, \text{atm} \times 0.112 \, \text{L}}{0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \times 358.15 \, \text{K}} \)
First, calculate numerator: \( 1.004 \times 0.112 \approx 0.1124 \)
Denominator: \( 0.0821 \times 358.15 \approx 29.40 \)
\( n \approx \frac{0.1124}{29.40} \approx 0.003823 \, \text{mol} \)

Step4: Calculate Molar Mass (\( M \))

\( M = \frac{m}{n} = \frac{0.565 \, \text{g}}{0.003823 \, \text{mol}} \approx 147.8 \, \text{g/mol} \). Rounding to three significant figures, \( M \approx 148 \, \text{g/mol} \). Wait, let's recalculate more accurately.

Wait, let's do the calculation with more precise steps:

\( M = \frac{mRT}{PV} \)

Plug in values:

\( m = 0.565 \, \text{g} \), \( R = 0.0821 \), \( T = 358.15 \, \text{K} \), \( P = 763/760 \, \text{atm} \), \( V = 0.112 \, \text{L} \)

\( M = \frac{0.565 \times 0.0821 \times 358.15}{(763/760) \times 0.112} \)

Calculate numerator: \( 0.565 \times 0.0821 = 0.0464 \); \( 0.0464 \times 358.15 \approx 16.62 \)

Denominator: \( (763/760) \approx 1.004 \); \( 1.004 \times 0.112 \approx 0.1124 \)

\( M = \frac{16.62}{0.1124} \approx 147.8 \approx 148 \, \text{g/mol} \)? Wait, maybe I made a unit conversion error. Wait, pressure: 763 mm Hg is very close to 1 atm (760 mm Hg), so \( P \approx 1.004 \, \text{atm} \), volume 0.112 L, T=358 K.

Wait, let's use \( R = 62.36 \, \frac{\text{L·mm Hg}}{\text{mol·K}} \) to avoid converting pressure to atm. Then:

\( M = \frac{mRT}{PV} \)

\( R = 62.36 \), \( P = 763 \, \text{mm Hg} \), \( V = 0.112 \, \text{L} \), \( T = 358.15 \, \text{K} \), \( m = 0.565 \, \text{g} \)

\( M = \frac{0.565 \times 62.36 \times 358.15}{763 \times 0.112} \)

Calculate numerator: \( 0.565 \times 62.36 = 35.23 \); \( 35.23 \times 358.15 \approx 12617 \)

Denominator: \( 763 \times 0.112 = 85.456 \)

\( M = \frac{12617}{85.456} \approx 147.6 \, \text{g/mol} \), which rounds to 148 g/mol? Wait, no, maybe I messed up the volume. Wait, 112 mL is 0.112 L, correct. Mass is 0.565 g. Let's check with R=0.0821:

\( n = \frac{PV}{RT} = \frac{(763/760) \times 0.112}{0.0821 \times 358.15} \)

\( 763/760 = 1.0039 \)

\( PV = 1.0039 \times 0.112 = 0.1124 \)

\( RT = 0.0821 \times 358.15 = 29.40 \)

\( n = 0.1124 / 29.40 = 0.003823 \, \text{mol} \)

\( M = 0.565 / 0.003823 = 147.8 \approx 148 \, \text{g/mol} \). Wait, but maybe the correct answer is around 148? Wait, let's recalculate:

Wait, 0.565 g / n, where n is (P*V)/(R*T). Let's use P in mm Hg, R=62.36 L·mm Hg/(mol·K):

n = (763 * 0.112) / (62.36 * 358.15)

763*0.112 = 85.456

62.36*358.15 = 62.36*358 + 62.36*0.15 ≈ 62.36*358 = 62.36*(300+58) = 18708 + 3616.88 = 22324.88 + 9.354 = 22334.234

n = 85.456 / 22334.234 ≈ 0.003826 mol

M = 0.565 / 0.003826 ≈ 147.7 g/mol ≈ 148 g/mol (three significant figures). Wait, but maybe I made a mistake in temperature? 85 + 273 = 358 K, correct. Volume 112 mL = 0.112 L, correct. Mass 0.565 g, correct. So the molar mass is approximately 148 g/mol? Wait, no, let's check with another approach. Wait, maybe the gas is something else, but the calculation gives around 148. Wait, maybe I miscalculated. Let's do it again:

\( M = \frac{mRT}{PV} \)

\( m = 0.565 \, \text{g} \)

\( R = 0.0821 \, \text{L·atm/(mol·K)} \)

\( T = 85 + 273.15 = 358.15 \, \text{K} \)

\( P = 763 \, \text{mm Hg} = 763 / 760 \, \text{atm} \approx 1.0039 \, \text{atm} \)

\( V = 112 \, \text{mL} = 0.112 \, \text{L} \)

Plug in:

\( M = \frac{0.565 \times 0.0821 \times 358.15}{1.0039 \times 0.112} \)

Calculate numerator: 0.565 * 0.0821 = 0.0464; 0.0464 * 358.15 = 0.0464 * 358 + 0.0464 * 0.15 = 16.6112 + 0.00696 = 16.61816

Denominator: 1.0039 * 0.112 = 0.1124368

\( M = 16.61816 / 0.1124368 ≈ 147.8 \, \text{g/mol} ≈ 148 \, \text{g/mol} \) (three significant figures).

Answer:

\boxed{148} (Wait, but maybe the correct answer is 148? Or maybe I made a mistake. Wait, let's check with exact values:

Wait, 0.565 g, V=0.112 L, T=358.15 K, P=763/760 atm.

\( n = \frac{PV}{RT} = \frac{(763/760) 0.112}{0.0821 358.15} \)

Calculate (763/760) = 1.003947

1.003947 * 0.112 = 0.112442

0.0821 * 358.15 = 29.4041

n = 0.112442 / 29.4041 ≈ 0.003824 mol

M = 0.565 / 0.003824 ≈ 147.75 ≈ 148 g/mol (three significant figures). So the answer is 148 g/mol.