Question

Nitrogen contain sample of urea 42.5% by mass% purity of urea in the sample is

Answer 1

Urea NH2CONH2

Molecular mass =  60
Mass of nitrogen = 28

Percentage of nitrogen in 100% urea = 100 x 28/60 = 46.67

Actual percentage = 42.5

Purity = 100 x 42.5/46.67 = 91%

Answer 2

Answer : The % purity of urea in the sample is, 91.2 %

Solution : Given,

Molar mass of nitrogen = 14 g/mole

Molar mass of urea = 60 g/mole

First we have to calculate the mass % of nitrogen in the sample of urea.

$\text{Mass }\% \text{ of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of urea}}\times 100$

As we know that the formula of urea is, $NH_2CONH_2$.

In the urea, there are 2 number of nitrogen atom. So, the mass of nitrogen will be, $2\times 14=28g$

Now put all the values in the above formula we, get

$\text{Mass }\% \text{ of nitrogen}=\frac{28}{60}\times 100=46.6\%$

Now we have to calculate the % purity of urea in the sample.

$\% \text{ purity of sample}=\frac{\text{Actual}\% \text{ purity of nitrogen}}{\text{Theoretical}\% \text{ purity of nitrogen}}\times 100=\frac{42.5}{46.6}\times 100=91.2\%$

Therefore, the % purity of urea in the sample is, 91.2 %