Question

A sample of Urea $\left(C{H}_4{N}_{2}O\right)$ has $42.5\%$ N. What is the percentage purity of the sample?

• A. 92.07
• B. 93.07
• C. 91.07
• D. 100

Answer 1

The correct option is C

91.07

We have Urea $\left(C{H}_4{N}_{2}O\right)$

Its molecular weight is 60

Nitrogen present in Urea is 28

Therefore percentage of nitrogen in urea is $(\frac{28}{60}\times 100)$ = $46.67\%$

Nitrogen in given urea is $42.5\%$. Therefore its purity is given by $(\frac{42.5}{46.67}\times 100)$ = $91.07\%$