Question

Question #0fbc0

Answer 1

The solution will have a molality of `5.55 * 10^(-4)"molal"`.

Explanation:

Molality is defined as moles of solute, in your case urea, divided by kilograms of solvent, in your case water.

Notice that you were not given the mass of water, which means that you must use the given volume and water's density to determine how much solvent you have.

STP conditions correspond to a temperature of 273.15 K, or `0^@"C"`. At this temperature, water has a density of

`rho_"water STP" = "0.9998425 g/cm"""^3`

Since the volume given to you is expressed in `"dm"""^3`, and you need the mass of water to be expressed in kilograms, convert water's density to `"kg/dm"""^3`.

`0.9998425cancel("g")/cancel("cm"""^3) * (1000cancel("cm"""^3))/("1 dm"""^3) * "1 kg"/(1000cancel("g")) = "0.9998425 kg/dm"""^3`

This means that the mass of the solvent will be

`0.3000cancel("dm"""^3) * "0.9998425 kg"/(1cancel("dm"""^3)) = "0.29995 kg"`

To determine how many moles of urea you have, use its molar mass

`0.0100cancel("g") * "1 mole urea"/(60.055cancel("g")) = 1.665 * 10^(-4)"moles urea"`

Therefore, the solution's molality is

`b = (1.665 * 10^(-4)"moles")/("0.29995 kg") = color(green)(5.55 * 10^(-4)"molal")`

The answer is rounded to three sig figs, the number of sig figs you gave for the mass of urea.