How many grams of water would you add to 1.00 kg of 1.48 m CH3OH(aq) to reduce the molality to 1.00 m CH3OH?
Ok...I got 480, and the systems says there's a rounding error. I did it again and I got 480. Someone help me?
I think 480 grams is the correctnumber; my best guess is that the problem is one of significant figures. I assume you are keying the answer into an on-line data base; try keying in 4.8 x 10^2 grams. Let me know if this works or not (either way).
nope. have to contact my teacher about this.
To find the number of grams of water to add, we need to start by calculating the initial number of moles of CH3OH in the solution:
1. Calculate the initial moles of CH3OH:
Given:
Mass of CH3OH = 1.00 kg
Molar mass of CH3OH = 32.04 g/mol (Carbon: 12.01 g/mol x 1 + Hydrogen: 1.008 g/mol x 4 + Oxygen: 16.00 g/mol + 1 = 32.04 g/mol)
Number of moles of CH3OH = (Mass of CH3OH) / (Molar mass of CH3OH)
= (1000 g) / (32.04 g/mol)
= 31.20 mol
Next, we need to calculate the initial amount of solvent in moles.
2. Calculate the initial moles of solvent (water):
Since the solution initially contains only CH3OH, the moles of water are equal to zero because no water is present. Therefore, the molality (moles of solute per kilogram of solvent) is equal to the molarity (moles of solute per liter of solution).
So, the initial molality of CH3OH is already 1.48 m.
To reduce the molality to 1.00 m, we need to determine the new amount of solvent required to dilute the solution.
3. Calculate the new amount of solvent (water):
Let's assume that we need to add 'x' grams of water.
Number of moles of water = (Mass of water) / (Molar mass of water)
= x / (18.02 g/mol) (Molar mass of water is 18.02 g/mol)
Since we want to achieve a molality of 1.00 m, we can set up the equation as:
(31.20 mol) / (1000 g + x g) = 1.00 m
To solve for 'x', we rearrange the equation:
x = (31.20 mol) / (1.00 m) - 1000 g
= (31.20 mol) / (1.00 mol/kg) - 1000 g
= 31.20 kg - 1000 g
= 31200 g - 1000 g
= 30200 g
Therefore, to reduce the molality to 1.00 m CH3OH, you need to add approximately 30200 grams of water, instead of 480 grams.
Note: The discrepancy in your previous calculation may have been due to rounding errors or using incorrect values in the calculations. Make sure to use precise values and pay attention to significant figures in each step.