My solution was this.
Assume 1 mol of sol'n, then:\[1 mol(sol'n)=0.9245mol(H_2O)+0.0755mol(urea)\]Then I found the ratio of mols of water to mols of urea:\[\frac{0.9245}{0.0755}=12.245\]
Next, I used this expression:\[mass_{sol'n}=mass_{H_2O}+mass_{urea}\]And since...\[mass_{compound}=MolarMass_{compound}*mols_{compound}\]You can do some substitution to change the above equation (mass sol'n...) into:\[mass_{sol'n}=(mols_{H_2O}*MM_{H_2O})+(mols_{urea}*MM_{urea})\]
We know that for this solution there are 12.245 times as many mols of water as there are of urea, so...\[1mol(urea)=12.245mol(H_2O)\]
Now, let's express this ratio in terms of x within the mass of solution equation:
\[mass_{sol'n}=(12.245)(x)*(MM_{H_2O})+(1)(x)*(MM_{urea})\]Now, let's plug in some numbers and see what we get...\[51.8g=(12.245)(x)*(18.016g*mol^{-1})+(x)*(60.062g*mol^{-1})\]After some algebra, we can find that...\[x=0.1846\]
If we look at the original equation, we can see that x is the same thing as the number of mols of urea, so we know that there are 0.1845mol urea. From here, you can convert this to grams.
After you convert it to grams, you can check your answer by checking to see that the molefraction is 7.55 x 10^-2, which indeed it is.