a sample of vapor weighing 0.745 g filled a 250 ml flask when placed in a water bath at 98 degrees celsius. the barometric pressure that day was 753 torr. calculate the volume of the vapor at stp, and the weight of one mol of the volitile liquid

To calculate the volume of the vapor at STP (Standard Temperature and Pressure) and the weight of one mole of the volatile liquid, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure in atmospheres
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin

First, we need to convert some of the values given in the problem:

1. Convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 98°C + 273.15
T(K) = 371.15 K

2. Convert the volume from milliliters to liters:
V(L) = V(mL) / 1000
V(L) = 250 mL / 1000
V(L) = 0.25 L

3. Convert the pressure from torr to atmospheres:
P(atm) = P(torr) / 760
P(atm) = 753 torr / 760
P(atm) = 0.990 atm

Now we can rearrange the ideal gas law equation to solve for the number of moles (n):

n = PV / RT

Substituting the values we have:

n = (0.990 atm) * (0.25 L) / [(0.0821 L·atm/(mol·K)) * (371.15 K)]
n ≈ 0.00991 moles

Next, we can calculate the molar mass of the volatile liquid and its weight:

Molar mass = Weight / Moles

Given that the weight of the sample of vapor is 0.745 g, we have:

Molar mass = 0.745 g / 0.00991 moles
Molar mass ≈ 75.1 g/mol

Therefore, the weight of one mole of the volatile liquid is approximately 75.1 grams.