Question

b. the molar mass of 0.250 g of gas that occupies 100 ml at a pressure of 160 kpa and a temperature of 22 °c is
1.63×10³g/mol
6.52×10³g/mol
0.0261g/mol
38.3g/mol

Explanation:

Step1: Convert units

First, convert volume $V = 100\ mL=0.1\ L$, pressure $P = 160\ kPa = 1.6\ atm$ (since $1\ atm = 101.325\ kPa$), temperature $T=(22 + 273)K=295K$.

Step2: Use the ideal - gas law $PV = nRT$

We know $R = 0.0821\ L\cdot atm/(mol\cdot K)$. From $PV=nRT$, we can find the number of moles $n$. Rearranging for $n$ gives $n=\frac{PV}{RT}$. Substitute the values: $n=\frac{1.6\ atm\times0.1\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times295K}$.
\[n=\frac{1.6\times0.1}{0.0821\times295}mol\approx6.52\times 10^{-3}mol\]

Step3: Calculate molar mass

The mass of the gas $m = 0.250\ g$. Molar mass $M=\frac{m}{n}$. Substitute $m = 0.250\ g$ and $n = 6.52\times 10^{-3}mol$ into the formula: $M=\frac{0.250\ g}{6.52\times 10^{-3}mol}\approx38.3\ g/mol$.

Answer:

38.3 g/mol (corresponding to the last option)