Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1 °C. What is the molar mass of lauryl alcohol from these data? See Table 13.3 for the normal freezing point and 𝐾_f of benzene.

Video transcript

Hello everyone today. We are being asked to solve the following problem. It's a solution composed of three g of D. H. A. In 30.2 kg of benzene freezes at 5.27 degrees Celsius. And we're supposed to calculate the molar mass of D. H. A. So the first thing I wanna do is we want to calculate the freezing point depression. We do that by taking The freezing point of Benzene which is 5.5°C. And what we're gonna do is we're going to take the number subjected by the 5.27 That we see in this question stem to give us 0.23°C. We're then going to take our freezing point depression. We're going to solve for our morality. And so we already have our .23°C and the constant or the freezing point malakal constant for Benzene is 5.07 degrees of Celsius morality. We're gonna multiply that by em. When we saw for em we're going to get the m equals 0.4536 morality morality is simply moles per kilogram. And so what we're gonna do is we're gonna take that 0.4536. And we're going to say that that is molds, pour kilogram and then we're gonna equal that to our three g of D. H. A. That we have, we're gonna place this over the molar mass and we're going to see wine a bit. That's how we're gonna solve for a number of moles. We're gonna take this and we're going to divide that by 0. kg. Rearranging this equation to isolate the three g for mila moles, we get nine point we get 9.72 times 10 to the negative third moles Is equal to that three g per molar mass, isolating molar mass. We get that molar mass is equal to three g over the 9.72 times 10 to the negative third moles. Which, when simplified, gives us an answer of 331 g per mole or the molar mass. I hope this helped. And until next time.