Calculating the Mass of SoluteA to Lower Vapour Pressure of Water

To lower the vapour pressure of water, we need to add soluteA, whose molecular mass is 40 gm/mol, to 180 gm of pure water. Let's look at the steps involved in calculating the mass of soluteA.

Step 1: Calculate the Vapour Pressure of Pure Water

The first step is to calculate the vapour pressure of pure water. We can use the Antoine equation, which relates the vapour pressure (P) of a substance to its temperature (T) using the following formula:

log(P) = A - B/(T+C)

For water, the Antoine coefficients are A = 8.07131, B = 1730.63, and C = 233.426. At room temperature (25°C or 298K), the vapour pressure of pure water is:

log(P) = 8.07131 - 1730.63/(298+233.426) = 3.1699

P = 10^3.1699 = 23.76 mmHg

Step 2: Calculate the Vapour Pressure of the Solution

Next, we need to calculate the vapour pressure of the solution after adding soluteA. According to Raoult's law, the vapour pressure of a solution is proportional to the mole fraction of the solvent (water) in the solution:

P(soln) = X(water) * P(water)

If we assume that soluteA is non-volatile and does not affect the vapour pressure of water, then the mole fraction of water in the solution is:

X(water) = 180 g / (18 g/mol * 1000 g/kg) / (40 g/mol + 180 g / 18 g/mol) = 0.909

Therefore, the vapour pressure of the solution is:

P(soln) = 0.909 * 23.76 mmHg = 21.60 mmHg

Step 3: Calculate the Mass of SoluteA

Finally, we need to calculate the mass of soluteA required to lower the vapour pressure of water to 4/5 of its original value. This means that the vapour pressure of the solution should be:

P(soln) = 4/5 * P(water) = 4/5 * 23.76 mmHg = 19.01 mmHg

Using Raoult's law again, we can solve for the mole fraction of water in the solution:

X(water) = P(soln) / P(water) = 19.01 mmHg / 23.76 mmHg = 0.801

The mole fraction of soluteA is then:

X(soluteA) = 1 - X(water) = 0.199

The mass of soluteA required is:

mass(soluteA) = X(soluteA) * 180 g / (40 g/mol) = 0.895 g

Therefore, we should add 0.895 g of soluteA to 180 g of pure water to lower its vapour pressure to 4/5 of its original value.

Is the answer 2.5g?