Question

Calculate the molality of a 20.0 percent by weight aqueous solution of nh4cl. (molecular weight: nh4cl = 53.5)

Answer 1

Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.  
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.  
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.   
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.

Answer 2

Answer:

$4.67~m$

Explanation:

If we have a 20% by weight solution indicates that in 100 g of solution we have 20 g of $NH_4Cl$. So, in the 100 g of solution we will have 80 g of $H_2O$ (100-20= 80). If we remember the molality equation:

$m=\frac{mol~of~solute}{Kg~of~solvent}$

On this case the solute is $NH_4Cl$, so we have to convert from g to mol using the molar mass:

$20g~NH_4Cl\frac{1~mol~NH_4Cl}{53.5g~NH_4Cl}$

$0.373mol~NH_4Cl$

Then we have to calculate the Kg of solvent ($H_2O$), so:

$80~g~H_2O\frac{1~Kg~H_2O}{1000~g~H_2O}$

$0.08~Kg~H_2O$

Finally, we have to divide these two values:

$m=\frac{0.373mol~NH_4C}{0.08~Kg~H_2O}$

$4.67~m$