A small cube of lithium 1density = 0.535 g/cm32 measuring 1.0 mm on each edge is added to 0.500 L of water. The
following reaction occurs:
2 Li1s2 + 2 H2O1l2 ¡ 2 LiOH1aq2 + H21g2
What is the freezing point of the resulting solution, assuming that the reaction goes to completion?

Question

A small cube of lithium 1density = 0.535 g/cm32 measuring 1.0 mm on each edge is added to 0.500 L of water. The 
following reaction occurs: 
2 Li1s2 + 2 H2O1l2 ¡ 2 LiOH1aq2 + H21g2 
What is the freezing point of the resulting solution, assuming that the reaction goes to completion?

Accepted Answer

Hello everyone today. You've been given the following question and asked to solve for it. It says the severe sodium and it's giving us a density with the radius of . cm is added 2.3 L of water. The sodium reacts with water to yield us this chemical equation here we are then asked to calculate the freezing point of the solution. So the first thing we wanna do is we want to calculate the volume of the sphere. So the volume of a sphere is going to be 4/3 pi r cubed. We're going to plug in our values that we have. We have 4/3 and we have pie which is a constant. Our radius is 0.72 cm. And we have to cube that and this is going to yield us a volume of 1.56 3, cm cute. We then must find the mass of this sodium that we have. So our second step is to find the mass and so the mass of this sodium is going to be equal to the density or the volume that we just saw for which is 1.535 centimeters cubed. And we're gonna go ahead and multiply it by our density which is 0.97 g per centimeters cubed. Our units are going to cancel out and we are going to be left with 1.5166 g. We must then find the moles of sodium hydroxide Which we take the mass that we just sold for. So 1.5166 g of sodium. And we're going to go ahead and multiply that by the molar mass of sodium. So we're gonna put one mole of sodium is equal to about 23 g per the periodic table. Never gonna multiply by the multiple ratio. So we're gonna use our coefficients from the chemical equation of the question. We're going to say every two moles of sodium on the denominator Is going to yield us two moles of sodium hydroxide. Our units are going to cancel out And we are going to be left with 0. moles of sodium hydroxide. We can now solve for the morality of our sodium hydroxide. So R. M of sodium hydroxide is going to be the moles that we just calculated the 0.660 moles of sodium hydroxide over our mass that we have, which is 0.3 kg are mass of our solvent, which is water over here. From the question, stem the leaders is going to be equal to the number of kilograms. And so When we saw for this, we get a morality of 0.22 M. We can then finally plug this into our freezing point depression equation, which is going to be the freezing point depression is equal to I or the van Hoff factor times our moral constant for the freezing point of our sodium hydroxide. And we're going to multiply that by our morality. And so I was going to be equal to two. Since the Vantaa factor is just how much, how many ions our compound associates into sodium hydroxide associated to sodium and hydroxide ions. The constant is going to be equal to 1. Celsius, Para morality. And then our morality is 0.22 M. Once all of our units cancel out, we are going to be left with a final temperature of 0.8184°C. And finally, to calculate Our freezing point, we're going to take the freezing point of water, which is 0°C and subtracted by the calculated freezing point from our previous calculation, which will yield us negative 0.82°C. This is our final answer. I hope this helped. And until next time.

A small cube of lithium 1density = 0.535 g/cm32 measuring 1.0 mm on each edge is added to 0.500 L of water. The following reaction occurs: 2 Li1s2 + 2 H2O1l2 ¡ 2 LiOH1aq2 + H21g2 What is the freezing point of the resulting solution, assuming that the reaction goes to completion?

Verified Solution

Key Concepts

Density and Volume Calculations

Stoichiometry of Chemical Reactions

Freezing Point Depression